LIBRARY 

OF  THE 

UNIVERSITY  OF  CALIFORNIA. 


Class 


u 


MECHANICS 

OF 

INTERNAL  WORK 

(OR  WORK  OF  DEFORMATION) 

IN  ELASTIC  BODIES  AND  SYSTEMS  IN  EQUILIBRIUM 

INCLUDING 

THE  METHOD  OF  LEAST  WORK 


BY 


TRYING   P.  CHURCH,   C.E. 

Assoc.  AM.  Soc.  C.E. 

Professor  of  Applied  Mechanics  and  Hydraulics,  College  of  Civil  Engineering, 
Cornell  University 


FIRST    EDITION 

FIKST    THOUSAND 

•-•-    -          '     . 
Of    THE 

UNIVERSITY 

OF 

ILJJF 

NEW    YORK 

JOHN  WILEY   &   SONS 

LONDON:    CHAPMAN  &  HALL,   LIMITED 

1910 


Copyight,  1910 

BT 
IRVING    P.    CHURCH 


THE  SCIENTIFIC   PRESS 

IMMONO  AND  COMPANY 
BROOKLYN,  N.  Y. 


PREFACE 


AMONG  the  more  modern  methods  used  in  dealing  with  elastic  bodies 
in  the  Mechanics  of  Materials,  those  based  on  the  properties  of  the 
derivatives  of  the  work  of  deformation  (or  "internal  work"  as  it  will  be 
called  in  the  present  volume),  as  first  thoroughly  established  by  Cas- 
tigliano  *  in  1879,  and  including  the  " Method  of  Least  Work,"  have 
been  but  sparingly  presented  in  American  text-books;  and  even  in 
those  works  which  include  applications  of  these  methods,  such  proofs 
of  the  methods  themselves  as  may  be  given  are  generally  so  meagre  in 
detail  that  the  student  is  left  much  in  the  dark  as  to  the  precise  condi- 
tions which  must  be  observed  to  render  the  methods  applicable. 

It  has  therefore  seemed  desirable  to  the  writer  that  a  text-book  be 
prepared  which  should  deal  exclusively  with  the  "  Mechanics  of  Internal 
Work  in  Elastic  Bodies  in  Equilibrium"  and  embody  special  fullness  of 
explanation  and  illustration,  not  only  in  the  demonstrations  of  the 
methods  to  be  employed  but  also  in  the  detail  of  the  applications;  and 
thus  indicate  clearly  the  use,  scope,  and  limitations  of  these  valuable 
methods. 

The  following  pages  are  the  result  of  an  attempt  in  this  direction; 
and  it  is  hoped  that  the  book  may  prove  useful  not  only  to  students 
of  engineering,  but  in  some  degree  also  to  members  of  the  profession. 

In  this  connection  the  attention  of  students  who  read  German  is 
directed  to  Miiller-Breslau's  standard  work  on  "Recent  Methods  in 

*  Castigliano  wrote  in  French,  his  book  being  entitled  "  Theorie  de  I'Equili- 
bre  des  Systemes  Elastiques."  It  was  published  at  Turin  in  1879  and  has 
been  translated  into  German.  Prof.  William  Cain's  paper  on  "Determination 
of  the  Stresses  in  Elastic  Systems  by  the  Method  of  Least  Work"  in  the  Transac. 
Am.  Soc.  Civ.  Engs.,  April,  1891,  is  a  valuable  contribution  to  this  subject; 
as  also  Prof.  George  F.  Swain's  article  "  On  tJie  Application  of  the  Principle 
of  Virtual  Velocities  to  the  Determination  of  the  Deflection  and  Stresses  of 
Frames,"  in  the  Journal  of  the  Franklin  Institute  for  Feb.,  March,  and 
April,  1883. 

iii 


211757 


iV  PREFACE 

Strength  of  Materials,  etc.*"  Books  in  English  by  Hiroi  and  Martin 
are  referred  to  in  the  present  work  (see  p.  89);  and  also  articles  by 
Tilden,  Hudson,  and  Mensch  (see  pp.  51,  53,  and  93). 

The  abbreviation  M.  of  E.  will  be  used  in  referring  to  the  writer's 
"  Mechanics  of  Engineering." 

CORNELL  UNIVERSITY,  ITHACA,  N.  Y., 
October  20,  1910. 

*  Die  neueren  Methoden  der  Festigkeitslehre  und  der  Statik  der  Baukonstruk- 
tionen.  Leipzig,  1893  (zweite  vennehrte  und  verbesserte  Auflage). 


CONTENTS 


CHAPTER   I.     EXTERNAL  AND  INTERNAL  WORK  OF  ELASTIC 

STRUCTURES 

PAGES 

§§1-20.  Gradual  Application  of  Loads.  Structure  to  Consist  of 
Bars.  Successive  Introduction  of  Elastic  Properties. 

•  Limitation  to  Zero  Initial  Stresses.  Proof  that  External 

Work  Equals  Internal  Work '. 1-14 

CHAPTER  II.    THE  DERIVATIVES  OF  INTERNAL  WORK 

§§  21-25.  Derivative  of  Internal  Work  with  Respect  to  an 
External  Force  or  "Load."  Proofs  of  Castigliano's 
Theorem,  both  Analytical  and  Geometrical.  Application 
to  a  Frame.  FraenkePs  Formula.  Derivative  of  Internal 
Work  with  Respect  to  Moment  of  Couple 15-25 

CHAPTER    III.     THE  "THEOREM  OF  LEAST  WORK";   WITH  APPLICA- 
TIONS TO  SYSTEMS  OF  BARS  IN  EQUILIBRIUM  UNDER  LOADS 

§§  26-35.  Statically  Indeterminate  Structures.  Example  of 
Frame  of  Six  Bars,  One  of  Them  Redundant.  Proof  of 
the  "  Principle  of  Least  Work."  Frame  with  Two  Redund- 
ant Bars;  and  More  than  Two.  Theorem  when  a  Bar  is 
Originally  Too  Short,  or  Too  Long,  to  Fit  into  Place ;  and 
when  there  are  Several  Such  Bars.  Temperature  Stresses.  26-44 

CHAPTER   IV.     INTERNAL  WORK  IN  BEAIVIS  UNDER  FLEXURE. 
APPLICATIONS 

§§  36-56.  Internal  Work  of  a  Bent  Beam;  both  Straight,  and 
Curved  (of  Large  Radius  of  Curvature).  Examples  I  to 
VIII  of  Straight  Prismatic  Beams.  Straight  Horizontal 
Beams  of  Variable  Moment  of  Inertia.  Continuous  Girders; 
both  Prismatic  and  Tapering.  Derivative  of  Internal 
Work  of  Beam  with  Respect  to  a  Statically  Indeter- 
minate Internal  Moment.  Proof  of  the  Theorem  of 
Three  Moments.  Values  of  the  Two  Integrals  in  the 
Latter  Theorem  in  Particular  Cases.  Applications  of 
Theorem  to  Examples 45-77 


VI  CONTENTS 

CHAPTER  V.     COMPOSITE  SYSTEMS  AND  CURVED  BEAMS 

§§  57-73.  Straight  Beam  Supported  by  Three  Others.  Trussed 
Beam  or  Brake-beam.  Cambered  Trussed  Beam.  Exam- 
ples. Straight  Beam  Supported  by  Two  Piers  and  Parabolic 
Cable.  "  Double-knee  "  Beam.  Davit  of  Uniform  Section. 
Curved  Beams  of  Circular  Form  of  Axis.  Semicircular,  and 
Segmental,  Hinged  at  Ends,  also  Case  of  No  Hinges  or 
Fixed  Ends.  Simpson's  Rule  for  Approximate  Integra- 
tion. Curved  Davit  of  Variable  Section  Treated  by  the 
Aid  of  Simpson's  Rule 78-122 

APPENDIX 123 

INDEX 125 


MECHANICS  OF  INTERNAL  WORK 


CHAPTER  I 
EXTERNAL  AND  INTERNAL  WORK  OF  ELASTIC  STRUCTURES 

1.  Gradual  Application  of  Loads.    The  propositions  or  theo- 
rems of  this  and  the  two  following  chapters  have  to  do  with 
elastic  bodies,   or  structures,   undergoing  slight  deformations 
within  the  elastic  limit.     Certain  points  of  the  structure  are  sup- 
ported on   fixed  (and   inelastic)  bodies,  while  at  other  points 
forces  or  loads  are  applied,  each  increasing  in  value  gradually 
and   simultaneously,  and  proportionately  to  the  others,  until 
their  final  values  are  reached  at  the  same  instant.     During  this 
gradual  application  of  the  external  or  applied  forces,  gradual, 
but  slight,  changes  of  form  occur  in  the  elastic  body  or  struc- 
ture itself. 

2.  Structure  to  Consist  of  Bars.     For  present  purposes  the 
body  or  structure  will  be  conceived  to  be  composed  of  straight 
elastic  bars,  each  being  pivoted  (pin-connected)  to  the  neighbor- 
ing bars,  at  its  two  extremities    (and  nowhere  else);   all  these 
bars  lying  in  the  same  plane  (plane  of  the  paper).     All  external 
forces  or  loads  are  to  be  applied  at  joints;  that  is,  at  the  pins. 
Each  pin  is  perpendicular  to  the  plane  of  the  bars. 

3.  Bars  Considered  Elastic  in  Succession.    In  establishing 
the  theorems,  however,  it  will  be  convenient  to  consider  that 
the  elastic  character  of  the  body  is  introduced  in  successive 
stages;   that  is,  the  body  will  first  be  considered  to  be  entirely 
rigid  with  the  exception  of  one  bar,  all  the  remainder  of  the 
body  being  treated  as  a  stiff  plate  or  rigid  body.     Then,  portions 
of  this  plate  will  be  considered  to  be  composed  of  elastic  bars; 


2  MECHANICS   OF    INTERNAL    WORK  §  4 

and  so  on,  in  successive  cases,  until  a  proposition  is  established 
for  a  body  composed  entirely  of  straight  elastic  bars. 

4.  Necessary  Bars.     Redundant  Bars.     If  there  are  more 
bars  than  are  necessary  for  the  structure  to  hold  its   shape 
(that  is,  to  hold  it,  aside  from  the  small  elastic  deformation 
which  will  occur  when  loads  are  applied)  it  is  said  to  contain 
"  redundant  bars  ";    though  as  to  which  bars  are  redundant 
may  be  an  arbitrary  matter,  more  or  less.     Thus,  if  a  set  of 
five  bars  form  the  four  sides  and  one  diagonal  of  a  quadri- 
lateral, all  of  these  bars  are  necessary  that  the  framework  may 
hold  its  shape  and  be  capable  of  bearing  loads  applied  at  certain 
joints,  two  other  joints  being  supported;  but  if  a  bar  be  placed 

/7\^  ft  in  the  position  of  the  other  diagonal,  the  frame  contains  one 
redundant  bar.  Any  one  of  the  six  bars  could  be  treated  as 
the  redundant  bar  and  the  other  five  as  the  "  necessary  bars  "; 
since,  whichever  one  were  omitted,  the  other  five,  connecting 
the  original  four  vertices  or  joints,  would  constitute  a  frame 
capable  of  holding  its  shape. 

5.  As  a  First  General  Theorem  it  will  be  proved  that  in  a 
structure  of  elastic  bars  the  •"  work  "  of  the  applied  forces  or 
loads  is  equal  to  the  work  expended  on  stretching  or  com- 
pressing the  bars,  all  these  loads  being  gradually  applied.     Or, 
briefly, 

External  work  =  Internal  work (1) 

It  is  restricted  to  cases  where  all  the  bars  are  accurately 
fitted  to  each  other  before  loading  and  are  under  no  initial 
'stress;  that  is,  under  no  stress  when  the  application  of  the 
applied  loads  begins;  so  that  all  the  bars  begin  to  stretch  or  to 
shorten  at  the  same  instant  and  the  stresses  in  them  increase 
in  direct  proportion  to  their  final  values.  Also,  there  must  be 
no  change  of  temperature  during  the  application  of  the  applied 
forces. 

In  the  case  of  a  bar  under  compressive  stress,  though  designed 
by  a  "  long  column  "  formula,  its  amount  of  shortening  is 
supposed  to  be  proportional  to  the  stress  at  any  instant. 

Case  I.  The  Structure  Contains  only  One  Elastic  Bar. 
Fig.  1  represents  a  structure  in  which  the  plate  N  is  rigid 
while  the  bar  hb  is  elastic.  M  is  a  fixed  rigid  support,  the 


I  5  EXTERNAL  AND  INTERNAL  WORK  3 

straight  bar  hb  being  "  pin-connected  "  both  to  N  and  M. 
A  load  P  is  applied  to  the  plate  at  the  point  e.  Originally,  at 
the  instant  of  first  application  of  the  load,  the  point  b  was  in 
position  a,  e  was  at  d,  and  the  bar  hb  was  under  no  stress.  As 
a  result  of  the  gradual  *  application  of  the  load  P  the  bar 
stretches  and  finally  its  extremity  b  reaches  the  position  of 
the  figure  and  the  load  (force)  P  reaches  its  final  full  value  P. 
At  this  final  instant  there  is  a  certain  tensile  stress  in  the  elastic 
bar  which  we  shall  call  T.  There  is  also  a  reaction,  R,  at  the 
pin  m,  where  the  rigid  plate  is  pivoted  to  the  fixed  support  M. 
One  extremity,  h,  of  the  elastic  bar  is  also  pivoted  to  M.  We 
have,  therefore,  the  rigid  body  N  under  the  action  of  the  three 


bc=Ji 


/k 


FIG.  1. 


forces  P,  T,  and  R,  and  in  equilibrium.  During  the  gradual 
application  of  the  load  its  point  of  application  has  moved 
from  d  to  its  final  position  e. 

Let  a  perpendicular  now  be  let  fall  from  d  upon  the  (present) 
line  of  action,  / .  .  .  k,  of  the  force  P.  The  distance  fe  is  then 
the  projection  of  de  upon  the  line  /.  .  .  k.  Similarly,  let  fall 
the  perpendicular  ac  from  point  a  upon  line  bh',  be,  the  projec- 
tion of  ab  upon  line  b  .  .  .  h,  being  thus  obtained. 

Since  the  plate  N  is  a  rigid  body  in  equilibrium  under  the 
three  forces  T,  P,  and  R,  let  us  now  apply  to  it  the  principle 
of  Virtual  Velocities  (M.  of  E.,  p.  68)  conceiving  it  to  execute 
a  small  displacement  of  precisely  such  character  and  extent  as 

*  If  the  force  or  load  P  is  due  to  gravity,  this  gradual  application  may 
be  supposed  to  be  brought  about  by  the  progressive  pouring  of  sand  into  a 
pail,  for  instance. 


4  MECHANICS   OF   INTERNAL   WORK  §  6 

to  bring  points  b  and  e  to  their  original  positions  a  and  d, 
respectively;  that  is,  a  small  rotation  about  point  m.  It  should 
be  remembered  that  this  small  displacement  is  purely  imaginary, 
and  that  while  it  is  taking  place  the  values  of  the  three  forces 
P,  T,  and  R,  remain  .  constant.  For  this  displacement  the 
"  virtual  velocity  "  of  T  is  be  and  occurs  on  the  positive  side 
of  the  point  b  [that  is,  on  the  side  toward  which  the  force  T 
(arrow)  points],  while  the  "  virtual  velocity  "  ef  of  the  force  P 
falls  on  the  negative  side  for  that  force  (on  the  side  opposite 
from  the  direction  in  which  P  points);  the  virtual  velocity  of 
force  R  being  zero.  Giving,  therefore,  the  proper  sign  to  each 
"  virtual  moment/'  or  product,  with  zero  as  right-hand  member 
of  the  equation,  we  have  [from  the  £(Pdw)=0  of  p.  68,  M. 
ofE.], 

T.cb-P.ef+R*0  =  0  ......     (2) 

But  evidently  the  distance  cb  is  not  only  the  virtual  velocity 
of  the  force  T,  but_also  the  difference  (practically  such),  X, 
between  its  length,  lib,  under  stress  and  its  original  length,  ha, 
when  under  no  stress.  That  is,  A  is  the  elongation  of  the  bar. 
Hence  we  may  write 

Py=TX,      .......    (3) 

in  which  t/,  or  ef,  is  the  "  displacement  "  of  the  point  d  in  the 
direction  of  the  force  P  (i.e.,  the  projection  of  de  on  line  of  force 

P). 

6.  External  Work;  and  Internal  Work.  While  the  last 
result  is  true  as  it  stands,  as  a  result  of  the  "  principle  of  virtual 
velocities  "  applied  to  a  rigid  body  at  rest  (the  displacement 
mentioned  being  only  imagined  to  take  place),  if  we  divide  both 
members  by  2  we  obtain  the  form 


to  be  interpreted  as  follows:  P/2  is  the  average  value  of  the 
force  P  during  the  gradual  movement  of  point  d  to  its  final 
position  e  (this  force  having  increased  from  zero  to  P),  while 
T/2,  similarly,  is  the  average  stress  in  the  elastic  bar  during 
its  gradual  elongation  (the  initial  stress  having  been  zero). 
Also,  y  is  the  projection  upon  the  line  of  P  of  tho  "  path/'  de, 


§  7  EXTERNAL  AND  INTERNAL  WORK  5 

of  its  point  of  application;  and  similarly  cb,  or  the  elongation 
A,  is  the  projection,  upon  the  line  of  the  force  T,  of  the  "  path/' 
«6,  of  the  point  of  application  of  this  latter  force.  Conse- 

p 
quently  the  product  -^y  is  called  the  work  done  by  the  force  P 

during  its  gradual  application  (and  this  is  called  "  external 
work  ";  and  if  other  loads  were  acting  we  should  add  a  similar 

T 

term  for  each  one);  while  the  product  -~-X  is  called  the  "  work 

of  deformation"  or  work  done  in  changing  the  length  of  the 
elastic  bar  from  a  condition  of  no  stress  to  its  final  condition 
(and  the  sum  of  a  number  of  similar  terms  for  a  number  of 
bars  forming  an  elastic  structure  will  be  called  the  "  internal 

work")- 

7.  Case  II.  The  Structure  Contains  Four  Elastic  Bars. 

Fig.  2  shows  a  structure  consisting  of  a  rigid  plate  N,  and 
four  elastic  bars,  namely,  1,  2,  3,  and  4,  connecting  the  plate 


FIG.  2. 

to  a  fixed  rigid  body  M.  To  the  rigid  plate  are  applied  two 
loads  PI  and  P2  acting  (in  this  final  position  of  the  plate) 
at  points  I  and  n.  When  the  loads  were  first  applied  these  two 
points  were  at  A;  and  m,  respectively,  and  the  values  of  the 
loads  were  each  zero.  The  pivots  b  and  /  where  the  elastic 
bars  are  connected  were  originally  at  points  a  and  e,  these 
points  forming  with  points  k  and  m  the  same  rigid  configura- 
tion which  they  form  in  their  final  positions  (the  plate  N  being 
rigid).  The  four  bars  comprehend  one  redundant  bar,  since 


6  MECHANICS    OF    INTERNAL    WORK  §  7 

three  alone  would  have  been  sufficient  to  enable  the  whole 
structure  to  hold  its  shape  and  its  connection  with  body  M\ 
but  the  proposition  would  be  equally  true  on  the  supposition 
that  only  three  bars  were  present.  During  the  gradual  appli- 
cation of  the  two  loads  varying  simultaneously  in  value  from 
zero  up  (and  proportionally)  to  their  final  amounts,  P\  and  P2, 
the  point  a  has  moved  to  some  new  position  b  and,  likewise, 
e  to/,  k  to  I,  and  m  to  n;  while  the  stresses  in  the  four  bars, 
whatever  their  initial  values,  attain,  respectively,  certain  final 
values,  T7!,  T2,  T3)  and  T±.  The  figure,  then,  shows  the  rigid 
body  N  in  this  final  state  of  equilibrium  and  acted  on  by  the 
six  forces,  viz.,  T\,  T2,  TZ,  T±,  P\  and  P2\  and  to  this  body 
we  shall  apply  the  principle  of  virtual  velocities;  regarding,  as 
the  small  displacement,  which  is  imagined  to  occur,  the  very 
movement  which  would  bring  the  four  points  6,  /,  /,  and  n, 
back  to  their  original  positions,  a,  e,  k,  and  m.  If  from  a  we 
draw  the  perpendicular  ad  upon  the  line  of  bar  3  (prolonged), 
thus  obtaining  the  point  d,  and  also  a  perpendicular  from  a 
upon  the  line  of  bar  1,  obtaining  the  point  c,  we  note  that  bd 
is  the  virtual  velocity  of  T%,  and  also  the  shortening,  ^3,  of  bar  3; 
while  be  is  the  virtual  velocity  of  T\  and  also  the  amount  of 
elongation,  X\y  of  bar  1.  Similarly,  drawing  at  the  point  e 
the  perpendiculars  eh  and  eg  we  determine  fh,  which  is  at  the 
same  time  the  virtual  velocity  of  T4  and  also  the  shortening  A4 
of  bar  4  ;  and  fg  as  the  virtual  velocity  of  T2  and  the  elongation 
X2  of  bar  2.  At  each  of  the  points  k  and  m,  dropping  perpen- 
diculars upon  the  lines  of  action  of  PI  and  P2j  respectively, 
we  determine  the  virtual  velocity  vl(  =  yi)  of  PI  and  also  that, 
on,  or  1/2,  of  P2.  Forming,  then,  the  algebraic  sum  of  the 
products  of  each  of  the  six  forces  by  its  virtual  velocity,  repre- 
senting the  respective  elongations  or  shortenings  of  the  bars 
by  AI,  >b,  etc.,  and  writing  the  sum  equal  to  zero,  we  obtain: 

Pzyz-Q,     .     .     (5) 


or,  by  transposition, 

4.   ...    (6) 


At  this  stage  of  the  present  case  there  is  no  object  in  dividing 
each  term  of  this  equation  by  2  and  calling  each  term  a  certain 


§  8  EXTERNAL  AND  INTERNAL  WORK  7 

amount  of  work  done  by,  or  upon,  each  force  during  the  gradual 
application  of  the  loads  PI  and  P^  since  it  may  have  been 
necessary,  before  this  application,  in  putting  the  bars  together, 
to  strain  one  of  the  bars  (the  last  one  fitted;  that  is,  the 
"  redundant  "  bar)  into  the  position  over  the  pivots  at  its 
extremities;  which  operation  would  create  in  all  the  four  bars 
certain  initial  stresses,  different  from  the  values  which  the 
stresses  in  these  bars  will  have  attained  at  the  end  of  the 
gradual  application  of  the  loads;  therefore,  the  mean  value 
of  the  stress  in  bar  1  (for  instance)  during  the  gradual  applica- 
tion would  not  be  one-half  of  the  final  value,  T\.  Hence 
eq.  (6)  will  be  allowed  to  stand  in  its  present  form,  as  a  mere 
consequence  of  the  principle  of  virtual  velocities,  without 
bringing  into  play  any  idea  of  the  "  work  done  "  by  any  force 
during  the  gradual  application  of  the  loads. 

8.  Special  Case  of  the  Preceding.  No  Initial  Stress. 
Let  us  suppose  that  although  there  is  one  redundant  bar,  the 
last  bar  fitted  is  of  just  the  proper  length  to  fit  over  the  pivots 
provided  for  its  extremities  without  strain;  there  will  then  be 
no  initial  stress  in  the  bars ;  and  when  the  gradual  application 
of  the  loads  begins,  all  the  bars  will  begin  simultaneously  either 
to  lengthen  or  to  shorten,  and  the  stress  in  each  will  change 
gradually  from  its  initial  value  zero  to  its  final  value;  so  that 
the  mean  value  is  one-half  the  final.  In  such  a  case  all  the 
terms  in  the  right-hand  member  of  eq.  (6)  are  positive,  whether  a 
bar  is  subjected  to  a  lengthening,  or  to  a  shortening.  Bar  1, 
for  instance,  suffers  an  elongation,  so  that  the  force  T\  points 
to  the  left  and  the  virtual  velocity  cb  falls  on  the  same  side  of 
the  point  b  as  that  in  which  the  arrow  (or  force)  T\  is  directed, 
so  that  the  product  T\\\  is  positive  in  eq.  (5).  Now  con- 
sidering the  case  of  bar  4,  we  note  from  the  figure  that  it  is 
subjected  to  a  shortening,  and  hence  to  a  compressive  force, 
T±.  This  force,  as  an  action  on  body  N,  must  therefore  point 
to  the  right  in  the  figure;  where  we  also  note  that  the  virtual 
velocity  for  this  force,  namely,  hf,  falls  on  the  same  side  of  / 
as  the  arrow  denoting  the  force ;  so  that  here  again  the  product 
(viz.,  T4^4)  will  be  positive  in  eq.  (5);  which  is  the  same  result 
that  we  reached  for  the  bar  in  tension.  From  this  it  follows,, 
then,  that  all  the  products  of  the  form  TX  in  the  right-hand 


8  MECHANICS    OF    INTERNAL    WORK  §  10 

member  of  eq.  (6),  in  this  case  of  no  initial  stress  in  -the  bars, 
are  positive;  from  which  fact  a  very  important  relation  will 
result,  in  future  propositions.  As  to  the  left-hand  member  of 
eq.  (6),  the  products  of  the  form  Py  may  be  positive,  or  may 
bo  negative ;  according  to  the  positions  of  the  respective  applied 
forces,  or  loads,  PI,  P^  etc.,  since  the  movement  of  the  rigid 
body  N  during  the  gradual  application  of  these  loads  may  be 
such,  and  the  forces  may  be  so  directed,  that  the  projections, 
or  virtual  velocities,  1/1,  2/2,  etc.,  may,  or  may  not,  fall  on  the 
same  side  of  I  (or  n,  respectively)  as  the  arrow  showing  the 
direction  of  the  force. 

9.  No  Redundant    Bars;    then  All  Products  TA  Positive 
in  eq.  (6).     If  in  Case   II  there    has  been  no  redundant  bars 
connecting  the  rigid  bodies  M  and  N  (in  other  words,  if  no 
more  than  the  "  necessary  "  bars  had  been  used)  there  would  be 
no  initial  stresses  in  the  bars;  so  that,  as  the  loads  are  gradually 
applied,  each  stress  increases  gradually  from  a  zero  value  to  its 
final,  and  each  term  on  the  right  in  eq.  (6)  is  positive,  as  already 
shown  for  the  case  of  no  initial  stresses  with  redundant  bars. 

It  should  therefore  be  noted,  that  in  both  cases,  viz.,  of 
redundant  bars,  and  of  simply  necessary  bars,  the  average 
stress  is  not  equal  to  one-half  the  final  unless  the  initial  stresses 
are  each  zero  (which  is  always  so  for  the  case  of  simply  necessary 
bars,  but  not  necessarily  true  in  a  redundant  system). 

10.  Case  III.  The  Structure  Contains  Five  Elastic  Bars, 
No  One  of  which  is  Redundant.     (N.B.     The  absence  of  a 
redundant  bar  does  not  vitiate  the  truth  of  the  proposition. 
One  or  more  such  bars  might  be  introduced  without  changing 
the  nature  of  the  demonstration;   the  object  of  omitting  them 
is  to  reduce  the  amount  of  detail.     See  §  14.) 

Here,  as  before,  we  have  (see  Fig.  3)  a  rigid  plate  TV  con- 
nected to  a  rigid  support  M  by  a  number  of  elastic  bars,  so 
arranged  as  to  be  five  in  number  and  to  involve  no  redundant 
bars.  Of  the  five  joints,  or  pivots,  having  to  do  with  these 
bars,  p  and  q  remain  stationary  during  the  application  of  the 
loads  PI  and  PI,  while  the  other  three,  now  at  b,  f,  and  s,  were 
originally  at  a,  e,  and  r,  respectively,  when  the  gradual  appli- 
cation of  the  loads  PI  and  P%  began.  At  this  initial  instant 
each  of  the  five  bars  was  under  a  zero  stress. 


§12 


EXTERNAL  AND  INTERNAL  WORK 


9 


11.  It  will  be  noticed  from  the  figure  that  the  only  bars  the 
stresses  in  which  act  on  points  of  the  rigid  body  N  are  bars  3, 
4,  and  5,  and  that  in  the  special  case  shown  in  this  figure  bar  3 
has  been  shortened,  so  that  its  action  against  body  N  is  a  force 
T3  pointing  toward  the  right.  As  to  whether  bars  4  and  5 
are  in  tension  or  in  compression  does  not  directly  appear  in  the 
figure  since  the  movement  of  joint  b  would  not  be  thoroughly 
known  except  in  some  particular  numerical  case.  As  a  supposi- 
tion to  be  consistently  followed  when  joint  b  comes  up  for 
consideration,  we  shall  suppose  that  the  action  of  bar  4  upon 
.body  N  is  &  tension,  and  hence  directed  upward  at  point  /; 


FIG.  3. 


.•and  that  the  action  of  bar  5  upon  body  N  is  also  a  tension, 
.and  hence  directed  to  the  left  at  point  s. 

12.  As  before,  we  shall  apply  the  principle  of  virtual  veloc- 
ities to  the  rigid  body  N  in  equilibrium  under  the  five  forces 
T3,  T±,  T5,  PI,  and  P2;  imagining  by  way  of  a  small  displace- 
ment that  the  points  of  application  (s,  /,  Z,  and  n)  shift  back 
to  their  original  positions  (r,  e,  k,  and  m)  and  letting  fall  from 
each  of  these  last  points,  respectively,  the  proper  perpendiculars 
upon  the  lines  of  action  of  the  respective  forces.  The  five 
forces  concerned  remain  constant  during  this  shift.  In  this 
way  we  obtain  the  virtual  velocity  ts  of  775;  fh,  that  of  T3', 
zf,  that  of  T4;  vl,  that  of  PI;  and  wn,  that  of  P2.  As  before, 
fh,  being  the  elongation  of  bar  3,  will  be  called  ^3;  also  vl,  y; 
and  wn,  y%]  whereas,  for  the  present,  st  and  zf,  as  quantities 


10  MECHANICS   OF   INTERNAL   WORK  §  12 

(lengths),  will   be  written   st  and  zf.     Forming  the  various 
products,  then,  or  "  virtual  moments,  "  we  have 


If  now  we  consider  the  joint  or  pivot  at  b  as  a  separate 

"  free  body,"  we  find  the  system  of  concurrent  forces,  shown 

in  Fig.  4,  acting  on  the  pivot; 
and  there  have  been  reproduced 
in  this  figure,  in  their  proper 
relative  positions,  the  point  o, 
the  four  perpendiculars  ac,  av, 
ad,  and  au,  as  also  the  prolonga- 
tions bu  and  bv  of  the  two  bars 
4  and  2.  In  other  words,  we 
have  in  this  figure  the  respective 
virtual  velocities  be,  bd,  bu,  and 

bv,  of  the  four  stresses,  or  forces,  acting  on  the  pin  b. 

Now  conceiving,  as  a  basis  for  applying  "  virtual  velocities," 

that  the  pin  b  is  displaced  through  the  small  distance  ba,  and 

forming  the  proper  products,  we  derive 

'T4.uo-T5.:vb  =  0.    .     .     .     (8) 


It  is  now  to  be  noted,  that  the  projection  cb  is  equal  to  the 
elongation,  ^i,  or  bar  1,  and  similarly,  that  the  projection  db 
is  the  elongation,  X2,  or  bar  2,  so  that  eq.  (8)  may  be  rewritten 


2-T4.uo-T5.vo  =  0  .....     (9) 
We  note  again,  examining  both  Fig.  3  and  Fig.  4  that 

zf-ub=  A4  =  the  elongation  of  bar  4; 
and  st—  vb=  >*5  =  the  elongation  of  bar  5. 

These  substitutions  having  been  made  in  the  equation  resulting 
from  the  adding  of  (7)  and  (9)  we  obtain,  after  transposition, 

Piyi  +  P2y2  =  T^  +  T2X2  +  7^3  +  Tth  +  TBls.    .     (10) 

13.  Attention  is  now  again  called  to  the  fact  that  each 
term  of  the  form  TX  in  the  right-hand  member  of  eq.  (10)  is 
positive,  and  this  feature  will  always  occur,  in  systems  having 
no  redundant  bars,  f<^r  reasons  previously  stated  (§8);  whereas,, 


§  16  EXTERNAL  AND  INTERNAL  WORK  11 

in  the  left-hand  member,  where  we  have  to  do  only  with  the 
applied  forces  and  the  respective  projections  y\,  etc.,  some  of 
the  terms  may  be  negative;  in  accordance  with  the  special 
conditions  of  any  particular  case. 

14.  In  case  there  had  been  redundant  bars  in  this  case, 
eq.  (10)  would  have  resulted  in  the  general  form  as  already 
derived;     but  if  the  idea  of  work  were  to  be  introduced,  we 
should  have  to  remember  that  at  the  beginning  of  the  gradual 
application  of  the  loads   the    various  elastic  bars  might  be 
already  under  stress  (that  is,  there  might  be  "  initial  stresses  ") ; 

rp       m 

so  that  -~-j  -~-,  etc.,  would  not  necessarily  be  the  average  values 

of  the  stresses  in  bar  1,  bar  2,  etc.,  respectively, during  the  gradual 
application  of  the  loads.  Also  the  terms  of  the  form  TX  in  the 
right-hand  member  of  eq.  (10)  would  not  all  necessarily  be 
positive  (though  such  would  be  the  case  if  the  bars  were 
so  accurately  fitted  originally  that  there  are  no  initial 
stresses). 

15.  It  now  becomes  evident  from  the  nature  of  the  steps 
already  taken,  that  eq.  (10)   is  general  for  any  structure  of 
elastic  bars  which  become  slightly  changed  in  length  by  reason 
of  the  gradual  application  of  loads  PI,  etc.,  and  that  it  is  true 
whether    (there    being    redundant    bars)    there    were    already 
stresses  (" initial  stresses")  existing  in  some  or  all  of  the  bars, 
or  not;   so  long  as  each  symbol  T  denotes  the  stress  tho  ot>oao 
in  a  bar  at  the  instant  when  all  the  loading  is  in  place  and  the 
body  at  rest,  and  X  denotes  the  change  of  length  occurring  in 
each  bar  as  due  to  the  gradual  loading  of  the  structure  (and 
not  the  difference  between  its  final  length  and  that  which  it 
would  have  under  no  stress  whatever. 

16.  Limitation  to  Zero  Initial  Stresses.     X  in  Terms  of  T. 
We  are  now  going  to  limit,  very  specifically,  the  conditions  of 
our  problems  by  stipulating  that,  before  the  structure  is  loaded, 
all  redundant  bars  are  of  just  the  proper  length,  respectively,  to 
fit  over  their  pivots  without  stress  of  either  kind,  tension  or 
compression.     Under  these  circumstances,  the  stress  in  each  bar 
is  zero  before  loading  takes  place,  and  the  quantity  X  denotes 
the  difference  between  the  length  of  a  bar  under  its  final  stress 
T  and  its  length  under  no  stress.     That  is,  X  is  the  elongation 


12  MECHANICS    OF    INTERNAL   WORK  §  17 

due  to  T  and  is  proportional  to  T,  bearing  to  it  the  relation  (see 
p.  203,  M.  of  E.), 

Tl 


where  F  is  the  sectional  area  (square  inches,  say)  of  the  bar 
(which  is  understood  to  be  a  prism  in  form,  and  homogeneous  in 
material),  I  is  the  length,  and  E  is  the  modulus  of  elasticity 
(pounds  per  square  inch,  for  instance)  of  the  material. 

Hence,  under  these  special  conditions,  eq.  (10),  after  division 
of  both  members  by  2,  may  be  rewritten  in  the  form, 


or  [see  eq.  (11)], 


the  form  in  eq.  (13)  being  of  such  a  nature  as  to  embody  certain 
convenient  conceptions,  as  follows  : 

17.  "  External  Work."     If  we  assume  that  the  external 
forces,  or  "  loads,"  are  applied  to  the  elastic  structure  simul- 
taneously, each  increasing  very  gradually  but  in  proportion  to 
the  others,  from  a  value  of  zero  to  its  full  (or  "  final  ")  value  PI 
(or  P2,  etc.)  (this  could  be  done  practically  by  the  progressive 
pouring  of  sand  into  pails,  for  instance)  ;   their  points  of  appli- 
cation will  be  deflected  or  displaced  through  the  various  dis- 
tances i/j,  2/2,  etc.,  measured  in  the  direction  of  the  corresponding 
load  or  "external  force";   and  the  product  of  JPi  (which  is 
the  average  value  of  load  No.  1  during  this  motion)  by  the 
deflection  y\  (in  direction  of  force)  is  called  the  "work"  done 
by  this  force  during  the  gradual  loading  of  the  structure. 
Similarly,  \P^^  is  the  work  done  by  load  No.  2,  etc.     (Some 
of   these  products   may   be   negative.)     Hence   the   left-hand 
member  of  eq.  (13)  may  be  called  the  "  External  Work,"  or 
work  of  the  applied  loads. 

18.  Internal  Work  Equals  External  Work.     During  this 
gradual  and   progressive   application   of   the   external   forces, 
tensile  and  compressive  stresses  are  created  in  all  the  elastic 
bars,  the  stress  in  each  bar  increasing  gradually,  and  simul- 


' 


$19  EXTERNAL  AND  INTERNAL  WORK  13 

taneously  with  those  in  all  the  others,  from  a  value  of  zero 
to  its  full  "  final  value  "  T  (pounds,  for  instance)  which  it 
reaches  at  the  same  instant  that  the  stresses  in  the  other  bars 
reach  their  respective  "  final  "  stresses.  In  this  progressive 
increase  of  stress,  the  stress  in  any  bar  is  always  directly  pro- 
portional to  the  change  of  length,  and  hence  the  average  stress 
in  any  bar  (for  purpose  of  formulating  the  work  done  in  pro- 
ducing that  change  of  length)  is  one-half  of  its  final  value  T, 

T 

and  the  product  of  average  stress  by  elongation,  that  is,  -~-X, 

& 

is  called  the  work  done  in  changing  the  length  of  the  bar.  Now 
the  right-hand  member  of  eq.  (12)  is  seen  to  consist  of  the  sum 
of  all  such  products  for  all  the  bars  of  the  structure  and  is  called 
the  "Internal  Work"  of  the  structure  or  system  of  elastic  bars, 
and  will  be  denoted  by  the  symbol  [7;  that  is  [see  also  eq.  (13)], 
U  denotes 


•.  •**'"''  •  •  (14) 

or, 


and  is  also  called  the  "  work  of  (elastic)  deformation"     Eq.  (12) 
or  (13)  can  therefore  be  read, 

External  Work  =  Internal  Work.    .     .     .     (16) 

19.  Note.  It  must  be  repeated  that  in  case  there  are 
redundant  bars  these  must  be  of  just  such  original  lengths  as 
to  fit  into  their  places  without  stress  before  the  structure  is 
loaded;  or,  to  express  it  in  another  way,  the  "  necessary  bars  " 
having  been  assembled  so  as  to  give  to  the  structure  its  proper 
form,  these  necessary  bars  being  without  strain  since  no  loads 
are  as  yet  applied,  the  other  bars,  viz.,  the  redundant  bars,  are 
supposed  to  be  of  just  the  proper  lengths  to  fit  upon  their 
various  pivots  without  subjecting  themselves  or  the  necessary 
bars  to  any  stress.  Hence  when  the  loads  are  now  gradually 
applied  to  the  structure,  the  initial  value  of  the  stress  in  each 
bar  is  zero.  It  must  also  be  understood  that  no  change  of  tem- 
perature occurs  in  the  structure  during  the  application  of  the 
loads. 


14  MECHANICS    OF    INTERNAL    WORK  §  20 

20.  The  arbitrary  phrases  just  adopted  (internal  and  external 
work,  etc.)  are  convenient,  but  not  really  necessary,  since 
eqs.  (10)  and  (12)  are  mathematically  correct  for  the  quantities 
indicated  by  the  respective  symbols,  whether  any  names  are 
given  to  the  products  in  question,  or  not;  and  it  is  really  imma- 
terial (so  far  as  the  final  result  is  concerned),  in  conceiving 
the  loads  to  be  placed  on  the  structure,  whether  they  are  applied 
in  the  manner  indicated  (simultaneously)  or  whether  they  are 
applied  in  succession,  each  gradually;  the  final  result  is  the  same. 

For  brevity,  the  fraction  ^  „  ,  which  is  dependent  onlv  on 

r  \rj\ 

the  dimensions  and  the  material  of  bar  No.  1,  may  be  called 
Ci  (and  similarly  for  the  other  bars  we  write  C2,  C3,  C4,  etc.; 
respectively),  and  hence  eq.  (14)  may  be  written 


U  =  internal  work,  =  WlTl2  +  W2T22  +  etc.',       .     (17) 
or, 

U=2(1,CT2);       .     .....     ...     t     .  i    .     .  (17a) 

while  the  expression  for 

external  work  is        %P\y  1+^22/2  +  etc.  .     .     (18) 

Note  that  each  term  in  (17)  is  positive;   while  in  (18)  some 
terms  may  be  negative. 


21 


DERIVATIVES    OF    INTERNAL    WORK 


15 


CHAPTER  II 
THE  DERIVATIVES  OF  INTERNAL  WORK 

21.  Castigliano's  Theorem.  Derivative  of  Internal  Work 
with  Respect  to  an  External  Force,  or  "  Load."  Statement: 
In  any  structure  of  elastic  bars,  at  rest  (under  no  initial  stresses, 
with  or  without  redundant  6ars;*  supported  by  fixed  points,  or 
by  fixed  and  smooth  supporting  surfaces,  or  both)  the  "  deriva- 
tive "  (or  "first  differential  coefficient  ")  of  the  total  internal 
work  with  respect  to  any  one  of  the  external  forces  (Pi,  for  instance) 
is  equal  to  the  displacement  (yi)  of  the  point  of  application  of 
that  force,  in  the  direction  of  the  force  (that  is,  the  projection,  upon 
the  action  line  of  the  force,  of  the  actual  displacement  of  the 
point  of  application). 

That  is,  we  are  to  prove  that 


In  the  next  figure,  Fig.  5,  the  network  of  elastic  bars  of  which 
the  structure  is  really  composed  is  not  shown.     The  continuous 


FIG.  5. 

straight  line  OacN  represents  the  structure  (for  example,  a 
bridge  truss)  under  no  load  whatever,  while  the  curved  con- 
tinuous line  ObeN  represents  the  same  structure  after  the  gradual 
application  of  the  two  (vertical)  loads  PI  and  P2.  0  and  N 
represent  two  fixed  supports.  The  point  0  of  the  structure  is 
fixed,  but  the  point  (or  joint)  N  bears  against  a  smooth  surface 
*  Note  the  restriction  stated  in  §  19. 


16  MECHANICS   OF   INTERNAL   WORK  §  21 

of  the  fixed  support  at  that  end.  During  the  application  of 
the  two  loads  the  points  of  application  have  moved  to  final 
positions  b  and  e  respectively,  so  that  their  total  (vertical) 
displacements  at  this  stage  are  ab,  or  y\,  and  ce,  or  y2)  respect- 
ively. ObeN  will  be  the  new  position  of  the  truss,  and  the  system 
of  external  forces  acting  upon  the  truss  consists  of  the  two 
loads  PI  and  P2  and  the  two  reactions  P0  and  Pn.  Whatever 
the  value  of  the  internal  work  may  be  at  this  stage,  we  know 
it  to  be  equal  to  the  external  work. 

Note.  It  is  to  be  noted  that  the  external  work  done  by 
the  reaction  PQ  is  zero,  because  point  0  does  not  move;  and 
that  the  work  done  by  Pn  is  zero,  because  its  point  of  applica- 
tion moves  at  right  angles  to  the  action  line  of  Pn,  and  hence 
the  projection  of  its  displacement  upon  the  action  line  of  Pn 
is  zero.  The  reaction  of  a  smooth  surface  is  necessarily  in  a 
line  normal  to  that  surface.  We  may  therefore  write 

Piyi        P2?/2       PpXO       PnXO. 

[/=—  7>~+—  ;  ~  +  —  2  —     —  2  —  ;     '       •  '   ^      '' 
that  is 

U  =  lPiyi+lP2y2.         .......     (20) 

We  shall  now  note  the  effect,  in  producing  an  increment, 
dU,  in  the  value  of  U,  of  giving  to  one  of  the  external  forces 
(say,  PI)  an  increment,  dPi.  This  increase  in  PI  will  cause 
the  truss  to  descend  slightly  to  a  lower  position  (dotted) 
Ob'e'N;  creating  changes  dyi  and  dy2  in  both  the  displacements 
2/i  and  y2.  (Pz  remains  of  the  same  value  as  before.)  This 
will  cause  a  small  addition  or  increment, 


in  the  external  work,  and  consequently  an  equal  increment 
(dU)  in  the  internal  work;  whence,  neglecting  the  term  involv- 
ing the  product  of  two  differentials,  we  have 


.     ....     (21) 

But  we  can  express  dU  in  another  form,  as  follows  :  We  first 
conceive  the  truss  to  be  restored  to  the  unstrained  position 
OacN  (by  removing  all  loads),  and  then  consider  the  pressures 
(or  loads)  at  a  and  c  to  have  (Pi  +  dPi)  and  P2  as  their  respective 


§21  DERIVATIVES    OF    INTERNAL   WORK  17 

final  values,  and  to  be  applied  gradually  and  progressively 
(from  zero),  while  always  maintaining  to  each  other  the  con- 
stant ratio  subsisting  between  their  final  values.  When  these 
final  values  are  reached,  we  find  the  structure  in  the  position 
(dotted)  Ob'e'N',  and  the  whole  external  work  done  in  this 
operation  of  "  deformation  "  (that  is,  from  state  OacN  to  state 
Ob'SN)  is 

;.    ..     .     (22) 


i.e.,  neglecting  the  term  involving  the  product  of  two  differentials, 
and  rearranging  terms, 

W  =  [iPit/i  +  JP2y2]  +  i(<*Pi)y  i  +  IPidyi  +  iP*^.      (23) 

Now  the  difference,  U'  —  U,  must  be  equal  to  dU,  the  incre- 
ment accruing  to  the  value  of  U  as  occasioned  by  giving  to 
PI  an  increment  dP\\  hence,  from  eqs.  (20)  and  (23), 

dU  =  WPi)yi+^Pidyi  +  ^P2dy2.      .     .     .     (24) 

But,  from  eq.  (21),  \Pifyi>+$Ffy*~fyttT\  and  hence 
eq.  (24)  becomes 

dU=(dPi)yi;    or,    ^=7i,      ....     (25) 

that  is,  the  derivative  of  U  (total  internal  work)  with  respect  to 
any  one  load,  .or  external  force,  P,  is  equal  to  the  displacement 
(y)  in  the  direction  of  this  P  (i.e.,  the  projection  of  the  actual  dis- 
placement upon  the  line  of  P)  of  the  point  of  application  of  P. 

This  is  the  fundamental  theorem  or  principle  in  the  subject 
of  internal  and  external  work  in  elastic  structures,  Castigliano's 
Theorem. 


If    therefore    the    total    internal    work,    U=^>    or 


(7=£(|CT2),  of  the  elastic  structure  can  be  expressed  as  a 
function  of  one  of  the  external  forces  (say,  PI),  we  may  find 
the  displacement  y\  of  the  point  of  application  of  that  load, 
in  the  direction  of  the  force,  by  differentiating  the  quantity  U 
with  respect  to  PI. 

Although  in  the  foregoing  there  has  been,  besides  PI,  only 
one  other  load  (P2)  [which  retains  a  fixed  value  during  the 
demonstration]  the  result  would  evidently  have  been  the  same 


18  MECHANICS   OF    INTERNAL    WORK  §  21 

in  case  there  had  been  any  number  of  other  loads  besides  PI. 
(P2  and  other  similar  loads  are  independent  of  the  value  of 
PI,  so  that  the  derivative  of  one  of  them  with  respect  to  PI 
would  be  zero).  The  reactions  P0  and  Pn  of  Fig.  5  are,  of 

/7P 

course,  dependent  on  PI;  so  that  -775-,  for  instance,  would  not 

ai  i 

be  zero. 

It  is  to  be  particularly  noted  in  the  foregoing  demonstration 
that  the  mode  of  support  is  of  such  character  that  the  reaction 
at  a  supporting  point  is  that  of  a  fixed  point  or  is  offered  by  a 
smooth  and  unyielding  surface.  A  rough  supporting  surface, 
along  which  slipping  may  occur,  is  precluded.  An  external  force 
doing  zero  work  may  be  called  a  "  neutral  "  force. 

Although  for  simplicity  the  loads  or  external  forces  PI,  P2, 
etc.,  have  been  taken  in  a  vertical  position  any  oblique  position 
might  have  been  assumed  for  them  without  altering  the  rigor  of 
the  demonstration;  nor  need  the  line  ON  of  the  structure  be 
taken  in  a  horizontal  position  necessarily.  Also  it  is  understood 
that  the  support  furnished  by  the  supporting  bodies  is  com- 
pletely effective  whatever  change  of  value  may  occur  in  the 
single  load  which  has  been  selected  as  a  variable. 

2  la.  Theorem  for  Variable  Displacement.  Castigliano  has 
proved  a  second  theorem  which  asserts  that  if  the  displacement 
(y)  of  the  point  of  application  of  a  load  (P)  (in  its  own  direction) 
be  made  to  vary,  the  displacements  of  all  the  other  loads  remain- 
ing fixed  in  value,  then, 

dJL-P. 
2jT 

but  this  theorem  is  of  little  practical  utility  and  has  not  come 
into  prominence. 

22.  Another  Proof  (Partly  Geometrical)  of  Castigliano's 
Theorem.  In  Fig.  6  have  been  reproduced  the  various  posi- 
tions a,  c,  6,  e,  etc.,  of  the  points  of  application  of  the  two 
external  forces  or  loads,  PI  and  P2,  of  Fig.  5,  the  values  of 
the  displacements  y\  and  3/2  and  their  increments  dy\  and  dy% 
being  indicated.  Before  PI  received  its  increment  dP\,  the 
work  done  by  the  two  forces,  each  increasing  gradually  from 
zero  and  in  proportion  to  final  values  P\  and  P2,  is  represented 


§22 


CASTIGLIANO'S    THEOREM 


19 


in  Fig.  6  by  the  two  shaded  triangles  abh  and  cem,  bh  being 
perpendicular  to  ab  and  made  equal  (by  scale)  to  the  final 
value  PI.  Similarly  (on  same  scale)  em  is  equal  to  P2.  The 
areas  of  these  triangles,  therefore,  are  respectively  the  work 
items  \P\y\  and  \P^y^.  If  now  PI  receive  its  increment, 
dP\  (a  little  more  sand  being  poured  into  that  pail),  while  P2 
remains  unchanged,  the  additional  work  done  at  b  will  be 
represented  by  the  (shaded)  trapezoid  bhkb' ',  and  that  at  e 
by  the  (shaded)  rectangle  emne';  ik  being  equal  to  dP\.  In 
this  first  mode  of  loading  the  structure,  then  (the  loads  being 
(P+dPi)  and  P2),  the  external  work  is  represented  by  the 
shaded  areas  of  Fig.  6. 


Let  now  the  structure  be  unloaded  (pails  emptied),  and  then 
reloaded ;  care  being  taken,  in  this  reloading,  to  make  the  amount 
of  force  (amount  of  sand  in  pails),  at  each  instant  during  the 
whole  process,  proportional  to  Pi+dPi,  and  P2,  as  final  values; 
instead  of  PI  and  P2.  In  this  second  mode  of  loading,  with 
same  final  values  PI  -\-dP\  and  P2  as  in  the  first,  the  total  external 
work  done  is  represented  by  the  two  areas  of  the  two  triangles 
aW  and  cue'.  The  former  area  is  seen  to  be  in  excess  of  the 
shaded  parts  in  the  left  of  the  figure,  by  the  area  of  triangle 
ahk'}  while  the  latter  is  less  than  the  shaded  part  in  the  right 
of  the  figure,  by  the  area  of  triangle  cmn.  Now  this  excess 
(ahk)  on  the  left  must  be  equal  to  the  deficiency  (cnm)  on 
the  right;  since,  whichever  of  the  two  modes  of  loading  be 
conceived  to  take  place,  the  final  state  of  the  component  bars, 
and  hence  the  amount  of  internal  work,  C7',  are  the  same  in 
the  two  cases;  and  consequently  the  amounts  of  the  total 
external  work  are  equal  in  the  two  cases.  That  is,  area  of 


20 


MECHANICS    OF    INTERNAL    WORK 


23 


triangle  ahk  =  area  of  triangle  cnm.     Now  area  ahk=  -{-triangle 
aik  —triangle  aih—  triangle  hikm,  i.e., 


+dyi]-±Pi  .  dt/i- 


and 


.   (25o) 
.    (256) 


Equating  (25a)  and  (256),  we  have,  neglecting  products  of 
two  differentials, 

.    ...    (25c) 


But,  by  eq.  (21),  the  right-hand  member  of  eq.  (25c)  is 
equal  to  JcH7;  and  hence,  finally, 


dU 


(25) 


which  is  Castigliano's  Theorem. 

23.  Example  of  Application  of  Castigliano's  Theorem. 

The  structure  in  Fig.  7  contains  seven  bars,  three  horizontal 
and  four  at  45°.     Pins  o  and  n  rest  on  the  smooth  horizontal 


FIG.  7. 

surfaces  of  two  fixed  piers,  so  that  the  reactions  P0  and  Pn 
are  vertical.  The  three  vertical  forces  (loads)  PI,  P2,  and 
PS  are  not  equal  to  each  other,  necessarily. 

It  is  required  to  find  the  vertical  displacement  of  joint  a, 
where  the  load  PI  is  applied  (i.e.,  the  projection,  upon  line  of 
PI,  of  movement  of  a).  (As  to  whether  a  moves  horizontally, 
and  to  what  extent,  depends  on  the  horizontal  sliding  of  one  or 
both  of  the  pins  at  o  and  n. 


§23 


CASTIGLIANO  S   THEOREM 


21 


There  being  no  redundant  bars,  after  finding  the  values  of 
PO  and  Pn  by  ordinary  statics,  viz. : 


(26) 


we  find  by  ordinary  statics  the  values  of  the  seven  stresses  in 
the  respective  bars  (in  terms  of  the  given  loads  and  the  reac- 
tions). As  to  whether  any  such  stress  is  tensile  or  compressive, 
this  is  immaterial;  since  the  internal  work  due  to  each  one  is  of 
necessity  positive  (see  §  9)  .  The  following  values  are  obtained  : 


and 


(27) 


Since  we  now  are  to  consider  PI  as  variable  each  T  must  be 
expressed  as  a  function  of  the  variable  P\  and  of  the  con- 
stants P£  and  P3.  This  is  done  by  the  use  of  eqs.  (26);  and 
the  derivative  of  each  T  with  respect  to  PI  is  then  easily 
obtained  and  set  down  in  the  proper  column.  In  this  way  the 
following  values  are  obtained : 


dPi     4' 

and    ^        4     , 
dT3 


4    ' 


V2 


l^L!'x     P  . 

~      +  ~"+       3' 


i     4 


4 


(28) 


OF   THE 

UNIVERSITY 

OF 


22 


MECHANICS    OF    INTERNAL    WORK 


23 


Before  we  apply  eq.  (25)  (Castigliano's  Theorem)  to  finding 
the  vertical  displacement  y\  of  joint  a  we  should  note  that,  in 
general  [see  eqs.  (14),  (17),  (17a)],  since 


the  derivative  -75-  can  be  written  in  the  form. 


...,et,.     .     (29) 
[Substitution  from  eq.  (28)  in  this  general  form  gives  rise  to 
less  complication  than  in  (17).]     Hence,  since  -T-  =  2/i,  we  have 


f3p+P*+P3\ 

^p,+y+T; 


(30) 


for  the  displacement  of  the  point  (or  joint)  a,  where  PI  is 
applied,  in  a  vertical  direction  (Pi  being  vertical);  as  due  to 
three  loads  (i.e.,  neglecting  the  weights  of  the  bars)  with  this 
mode  of  support. 

23a.  FraenkePs  Formula.  Since  the  three  loads  PI,  P2,  and 
P3  may  have  any  values  so  long  as  the  elastic  limit  is  not  passed 
in  any  bar,  it  is  evident  that  by  making  P\=zero  we  obtain 
the  deflection  of  point  a  as  due  to  the  loads  P2  and  P3  only; 
which  leads  to  the  following  convenient  rule:  To  find  the  dis- 
placement, in  a  given  direction,  of  a  joint  (of  a  loaded  truss  or 
elastic  structure)  where  there  is  no  load  applied,  conceive  a  load 
or  force,  PI,  applied  at  this  joint,  in  the  given  direction,  form 
an  expression  for  yi,  by  the  same  procedure  as  shown  in  fore- 
going example;  then  make  PI  =  zero  (Fraerikd's  Formula). 
N.B.  If  any  of  the  "  loads,"  or  forces,  acting  on  the  truss  in 
Fig.  7  were  not  vertical,  either  pin  o  or  n  would  have  to  be 
fixed. 


§24 


CASTIGLIANO  S   THEOREM 


23 


24.  Numerical  Case  of  Foregoing  Example  (in  §  23).    Let 

the  lengths  of  the  bars  and  the  amounts  of  the  loads  be  as 
shown  in  Fig.  7a,  viz. : 

Z1=Z4  =  /6  =  20ft.  =  240in.; 
/2  =  Z3=Z5  =  Z7  =  14.14  ft.  =  169.7  in.; 
PI  =  8  tons,     P%  =  2  tons,     and     P3  =  4  tons. 

Let  the  sectional  area  of  each  bar  be  Fi  =  F2  =  F3,  etc., 
=  2  sq.in.,  and  the  value  of  E  the  same  for  all,  30,000,000  Ibs. 
per  sq.in. 

We  have,  therefore,  with  the  inch  and  ton  as  units, 

L  240 


2X15000 


0.008;     also 


while 


0.008; 


0.00565. 


Above  values  in  the  various  terms  of  eq.  (30)  give  rise  to 

L 20- J  2  tons  j 

I %- 20- H 

fe  1       *  SrZ  6      ?  mk: 


1» 

FIG.  7a. 

f  (0.008)     (6  + 1  + 1)  =  +  0.0480  inches 

|(0.00565)(6 +  1  +  1)  =+0.0678  " 

-1(0.00565)  (1-2  +  1)=  -0.0000  " 

1  (0.008)     (4  +  2  +  2)  =  +  0.0320  ' ' 

1(0.00565)  (2 +  1-1)=  +0.0056  " 

1(0.008)     (2 +  1+3)  =+0.0120  " 

i(0.00565)(2  +  l +3)=  +0.0169  " 

and   hence  the  vertical   dis- 
placement 


?/i=  +0.1823  inches. 


24  MECHANICS    OF    INTERNAL    WORK  §  25 

(N.B.  It  should  be  understood  that  in  this  case  the  pins  at 
the  two  supports  are  free  to  slide  horizontally  on  the  smooth 
horizontal  surfaces  of  the  fixed  piers,  where  the  reactions  are 
therefore  vertical  in  direction;  and  that  the  m  veinentsof  these 
pins  are  at  right  angles  to  the  reactions  so  that  the  external 
work  done  by  the  reactions  is  zero.  If  the  piers  could  yield 
or  settle  during  the  gradual  increase  of  the  reactions  (due  to 
gradual  application  of  loads)  this  would  not  be  true  (zero  work).) 

25.  Derivative  of  Internal  Work  with  Respect  to  the 
Moment  of  an  External  Couple.  Among  the  external  forces 
holding  an  elastic  structure  in  equilibrium  let  there  be  a  couple 
consisting  of  two  equal,  parallel,  but  oppositely  directed  forces, 
PI  and  P2,  applied  at  the  extremities  of  a  rigid  cross-bar  bd, 
attached  to  an  elastic  structure.  Let  a  denote  the  perpen- 
dicular distance  between  PI  and  P2  ("  arm  "  of  the  couple). 


During  the  gradual  application  of  the  loads  or  external 
forces  to  the  structure,  the  point  of  application  of  PI  has  been 
displaced  from  e  to  b  (the  projection  of  this  displacement  upon 
the  line  of  P\  may  be  denoted  by  y\) ;  that  of  P2,  from  c  to  d 
(its  projection  on  the  line  of  P2  being  y2).  Then,  by  eq.  (25), 
we  have 

dU  '      dU 

_=2/1.    and    _  =  2/2 (31) 

The  angle  <f>,  between  the  original  and  the  final  position  of 
the  rigid  cross-bar,  is  so  small  in  any  practical  case  that  for 
tan  <f>  we  may  write  <j>  (i.e.,  in  " radians"  or  circular  measure) 
and  note,  after  drawing  be'  parallel  to  ec,  that 


a          a  •'•' <32> 


$  25  EXTERNAL  COUPLE   THEOREM  25 

in  which  for  tan  </>  we  now  write  </>;    and,  for  y\  and  3/2,  their 
values  from  eq.  (31);  whence 

dU  dU 


But  a  .  dPi=d(Pia)  =the  differential  of  the  moment    M, 
=  (Pid)  (  =  P20>)  of  the  couple;   and  the  left-hand  member  of 
>e(|.  (33)  expresses  the  complete  derivative  of  U  with  respect  to 
M:  and  hence  we  may  write 


Evidently,  from  the  figure,  a  positive  value  for  6  refers  to 
a  "  clockwise  "  displacement,  or  turning,  of  the  cross-bar,  if 
the  couple  itself  has  a  clockwise  moment;  i.e.,  in  general,  a 
positive  </>  means  an  angular  displacement  of  the  same  nature 
as  that  of  the  moment  of  the  couple;  and  vice  versa. 

For  instance,  a  value  of  0.0174  for  <j>  means  an  angle  of  1°. 


26 


MECHANICS    OF    INTERNAL    WORK 


26 


CHAPTER  III 

THE  "  THEOREM  OF    LEAST  WORK";     WITH  APPLICATIONS  TO 
SYSTEMS  OF  BARS  IN  EQUILIBRIUM  UNDER  LOADS 

26.  Statically  Indeterminate  Structures.    A  loaded  struc- 
ture  containing  one  or  more  redundant  (or    "  unnecessary  ") 
bars  is  called  "statically  indeterminate";   since  the  stresses  in 
the  bars  cannot  be  determined  by  ordinary  statics.     For  their 
determination  we  must  have  recourse  to  the  elastic  properties 
of  the  bars  or  members.     Castigliano's  Theorem,  bringing  into 
play,  as  it  does,  the  displacement  of  any  joint  and  the  elastic 
change  of  form  or  "  deformation  "  of  each  bar,  may  be  availed 
of  to  prove  a  second  relation  or  theorem,  the  direct  outgrowth 
of  the  first,  called  the  "  Theorem  of  Least  Work,"  the  application 
of  which  to  practical  cases  has  the  great  advantage  of  involving 
very  simple  and  direct  procedures. 

As  a  preliminary  step  in  the  development  of  this  second 
theorem  it  will  be  useful  to  take  up  the  study  and  solution  of  a 
simple  case,  from  which  the  general  form  of  the  relation  is 
readily  inferred ;  as  follows : 

27.  Statically   Indeterminate    Frame    of   Six   Bars.     In 

Fig.  9  is  shown  a  frame  of  six 
bars,  in  a  vertical  plane,  forming 
the  sides  of  a  square  and  both 
diagonals,  with  one  (vertical) 
load,  PI,  as  shown,  suspended 
on  the  lowest  pin  at  e.  Evi- 
dently there  is  one  redundant 
bar  (bars  1  and  2  are  not  con- 
nected in  any  manner  at  A). 

The  pins  at  o  and  n  are  sup- 
ported on  the  horizontal  smooth 
surfaces  of  the  fixed  supports 

(at  the  same  level),  the  reac- 

P1 
tions  from  which  are  therefore  vertical  and  each  equal  to  --r 


§  28  STATICALLY    INDETERMINATE    FRAME  27 

since  the  only  load,  PI,  is  midway  between.  [Or,  pin  o  might 
be  fixed,  and  pin  n  free  to  slide;  the  reactions  would  still  be 
vertical  and  each  would  be  JPi.] 

It  is  assumed,  as  before  stipulated  (§  19),  that  all  of  the 
six  bars  are  of  just  the  proper  lengths  to  enable  them  to  be 
fitted  together  without  creating  stress  in  any  bar.  (For 
example,  if  all  the  bars  except  bar  1  had  been  put  together 
and  bar  1  were  too  short  or  too  long  for  its  extremities  to  fit 
over  the  pins  at  a  and  e,  it  could  not  be  placed  in  position 
without  a  forcible  stretching  or  shortening,  and  when  this  had 
been  accomplished  stresses  would  be  created  in  all  the  bars 
("  initial  stresses  ").  Such  a  condition  is  specially  precluded 
here.  The  load  P\  being  now  gradually  applied,  it  is  required 
to  determine  the  final  stress  T\  (tension,  evidently)  in  bar  1; 
and  ultimately  T2,  T3)  etc. 

The  method  of  solution  to  be  used  depends  on  the  geomet- 
rical fact  that  the  amount  of  elongation  produced  in  the 
(vertical)  bar  1  must  be  equal  to  the  difference  between  the 
(downward)  vertical  displacements  of  its  extremities,  i.e.,  of 
joints  a  and  e]  since  if  these  displacements  were  equal  it  would 
imply  that  no  change  had  occurred  in  the  length  of  bar  1.  We 
shall  therefore  find  an  expression  for  the  downward  vertical  dis- 
placement 2/0  of  joint  a  in  terms  of  the  unknown  T\,  using  the 
method  of  §§  21  and  23  (Castigliano's  Theorem);  and,  similarly, 
an  expression  for  the  downward  vertical  displacement,  yej 
of  joint  e  in  terms  of  T\]  and  then  subtract  the  former  from 

]   rp 

the  latter  and  place  this  difference  equal  to   J  J  .  or  CiT\. 

iSifi 

which  expresses  the  elongation  AI  of  bar  1  in  terms  of  T\. 
The  resulting  equation, 

y.-y  <*  =  €&!,      • (35) 

will  contain  but  one  unknown  quantity,  viz.,  TI,  which  is 
then  easily  determined. 

28.  Detail  of  Solution.  Conceive  bar  1  to  be  removed, 
but  that  its  influence  on  the  remainder  of  the  loaded  structure 
is  fully  and  completely  maintained  by  the  vertical  down- 
ward force  TI  (unknown)  applied  at  a  and  an  equal  vertical 
upward  force  T\  applied  at  e.  For  clearness,  however,  in 


28 


MECHANICS    OF    INTERNAL    WORK 


§28 


subsequent  mathematical  work,  let  us  denote  the  T\  at  a  by  T, 
and  that  at  e  by  T'.    We  now  have  Fig.  10  in  place  of  Fig. 


9  and  note  that  it  shows  a  struc- 
ture containing  no  redundant  bars ; 
T  and  T'  playing  the  part  of  (un- 
known) external  forces,  like  PI 
and  the  two  reactions;  each  of 
these  reactions  being  now  ex- 
pressed in  the  form 


4(P+T-T') 


•J-(p+T-T') 


Now  the  downward  vertical  dis- 
placement    of     a     (ya,    say)     is 

dU 
ya==——')  the  external  force  T  being  vertical  and  downward,  so 

that  the  desired  ya  is  the  projection,  on  the  line  of  T,  of  the 
actual  displacement  of  joint,  or  pin,  a. 

The  bars  of  our  present  structure  being  2  to  6  inclusive, 
we  have  [see  eq.  (29)], 


Similarly,  the  displacement  of  pin  e  in  the  direction  of  the 
external  force  Tf  is  given  by  the  relation 


(37) 


But  since  the  force   Tf  points  vertically  upward  and   ye 
denotes  the  downward  vertical  displacement  of  pin  e,  we  have 


(38) 


From  ordinary  statics  we  obtain  the  values  of  T2,  T%,  etc., 
in  terms  of  T  and  T'  and  of  the  load  PI,  as  shown  in  the  follow- 
ing table  (in  which  the  second  and  third  columns,  containing 
certain  derivatives  to  be  used  later,  have  been  obtained  by 
obvious  means) : 


28 


STATICALLY    INDETERMINATE    FRAME 


29 


T    Pt- 


\/2 


dT4_   I 
dT  ~ 


dT,       1 

dT     V2 


— 8  =  0 
dT 


dT' 


If  now  we  write  T'  equal  to  T7  and  each  equal  to  T\,  and 
substitute  in  eqs.  (36)  and  (37)  the  values  just  given  in  the 
table,  we  obtain 


(39) 


and 


i-rj.j.    (40) 


Hence  the  value  of  ye—ya  will  be,  after  substituting  —  yf  for  y 
[see  eq.  (38)], 


and  this  should  =  AI,  i.e., 
is, 


=  elongation  of  bar  1).      That 


=0;.     .     (41) 


and  finally,  solving, 


2Ci  +  2C2  +  C3  +  C4  +  C5  +  CQ' 


(42) 


30  MECHANICS   OF    INTERNAL    WORK  §  28 

(For  instance,  if  the  coefficients  Ci,  C2,  C%,  etc.,  were  all 
equal  we  should  obtain  77i  =  f  PI.) 

But  we  shall  now  show  that  eq.  (41)  is  nothing  more  than 

what  may  be  obtained  by  writing  -77^-  =  0:     where   U  is  the 

al  i 

total  "  internal  work  "  of  the  whole  framework,  that  is,  of  all 
the  six  bars,  including  bar  1  itself. 

Since  U  =  JCi  T12  +  W2T22  +  .  .  .+iC67762,    what   we    under- 

IK    du  • 
stand  by  -       is 

dT« 

(43) 

Hence,  utilizing  the  table  following  eq.  (38),  where  we  find  T2, 
T3,  etc.,  as  functions  of  T  and  T  ',  each  of  which  will  now  be 


replaced  by  its  real  value,  TI,  so  that          =  "+3^7;  etcv 
we  find 


,  •     (44) 


V2    V    V2 
and  if  this  expression  be  written  equal  to  zero,  we  obtain 


-C6  •      ~^;      (45) 

which  is  identical  with  eq.  (41)   and  will  yield    the    correct 
value  of  TI. 

But  the  value  of  TI  obtained  from  eq.  (45)  is  a  value  which, 
from  the  principles  of  the  differential  calculus,  would  cause  U, 
the  total  internal  work  of  the  structure,  to  be  a  minimum; 
that  is,  of  all  values  that  could  be  proposed  for  TI,  that  one 
is  the  true  and  only  possible  one  which  makes  the  expression 
for  U  a  minimum.*  This  is  called  the  principle  of  least  work, 

*  Sine    dU/aT,=  0.     See  Note  C,  Appendix. 


§29 


PRINCIPAL    OF    LI  AST    WORK 


31 


\ 


a" 


first  established  by  Castigliano  in  1878 ;  and  it  will  now  be  proved 
in  its  general  form. 

29.  Principle  of  Least  Work.  General  Proof.  (For  a 
loaded  elastic  structure  containing  one  redundant  bar;  all 
bars  closely  fitted  originally  without  initial  stress.)  Select  the 
bar  to  be  considered  as  the  redundant  bar  and  let  T\  denote 
the  final  stress  in  it.  In  Fig.  11  ae  represents  this 
bar  before  the  structure  is  loaded.  As  a  result 
of  the  loading  the  joint  a  has  moved  to  a"  and 
the  other  extremity  of  the  bar,  viz.,  joint  e,  has 
moved  to  e" .  Perpendiculars  having  been  let  fall 
from  a"  and  e"  upon  the  original  position,  ae,  of 
the  bar  (when  unstrained),  we  note  that  ya,  =  aa', 
is  the  displacement  of  joint  a  in  the  direction  of 
the  bar,  and  that  yej  =  ee',  is  the 'displacement  of 
joint  e  in  the  direction  of  the  bar.  If  we  suppose 
the  bar  to  have  stretched  as  a  result  of  the  ap- 
plication of  the  loading  evidently  the  difference 

j   m 

ye  —ya  is  the  elongation,  =  AI,  =-=r^rj  =CiTi,  i.e., 


where  T\  is  the  final  stress  in  the  bar. 

Now  conceive  the  bar  to  be  removed,  the  forces  which  it 
exerts  on  the  joints  a  and  e  at  its  extremities  being  applied  at 
those  joints,  respectively,  to  preserve  the  remaining  bars  in  the 
same  position  and  state  of  stress  as  before.  These  forces  are 
equal,  and  act  in  the  same  line,  but  point  in  opposite  direc- 
tions. For  this  case  (bar  in  tension)  they  point  toward  each 
other  (see  Fig.  11)  and  will  be  called  T  and  T'  respectively, 
although  they  are  actually  equal,  each  being  equal  to  T\.  For 
the  remaining  bars  T  and  Tf  play  the  part  of  external  forces; 
i.e.,  either  T  or  Tf  might  be  conceived  to  vary,  i.e.,  change 
its  value,  the  other  remaining  constant,  without  destroying  the 
equilibrium  of  the  structure  formed  by  the  remaining  bars 
(note  that  such  a  statement  as  this  could  not  be  made  if  the 
ibar  in  question  were  not  a  redundant  bar). 

The  remaining  bars  constituting  a  statically  determinate 


32  MECHANICS    OF    INTERNAL    WORK  §  29' 

system,  then,  with  T  and  T'  playing  the  part  of  independent 
external  forces,  and  the  internal  work  of  this  system  being 

dU" 

denoted  by  17",  the  result  of  performing  the  operation  -^ 

would  be  an  expression  for  the  displacement  of  joint  a  in  the 
direction  of  T,  i.e.,  for  ya,  and  no  change  of  sign  is  necessary, 
since  joint  a  has  been  displaced  in  the  direction  in  which  T 

dU" 
points;   but  the  expression  given  by  the  operation  -^  would 

have  to  be  taken  with  a  contrary  sign  to  give  the  displacement 
ye,  since  joint  e  has  been  displaced  in  a  direction  (measured 
parallel  to  T')  opposite  to  that  in  which  T'  points.  That  is,, 
otherwise  expressed,  we  have 

dV"  dU" 

ya=~'  ye=  ~~''    '    '*'••    ' 


dT 
Hence  eq.  (46)  becomes 

dU"    dU"  dU"    dU 

~~dT==Cl   i;    Le''    ~d7r+ 


dU"    dU" 
Now  since  T  =  T'  =  Ti,  the  terms  "^+^7  are  the  same 

dU" 
thing  as  -r^r,  i.e.,  the  complete  derivative  of  V"  with  respect 

to  Ti,  U"  being   the    internal    work    of  all    the  bars  except 

d  [GiTi2~] 
bar.l;   while  C\T\  is  nothing  more  than  -JTJT}  —  ^  —    ,  i.e,    the 

derivative  of  the  internal  work  of  bar  1  itself,  with  respect  to 
TI.  Hence  the  left-hand  member  of  eq.  (48)  is  simply  the 
derivative,  with  respect  to  stress  T7!  in  bar  1,  of  the  internal 
work  of  all  the  bars,  including  bar  1,  i.e.,  of  the  total  internal 
work,  U,  of  the  original  complete  system  of  bars.  Eq.  (48) 
then,  takes  the  form 

dU 

dfT°'      '    '    .....     (49) 

as  the  condition  or  equation  the  solution  of  which  will  give  the 
value  of  the  stress  T\  in  bar  1. 


§  30  TWO  REDUNDANT  BARS  IN  FRAME  33- 

In  other  words,  of  all  values  that  could  be  proposed  for  Tt 
the  actual  or  true  value  is  the  one  which  would  make  the  func- 
tion U ' ,  or  total  internal  work,  a  minimum.  Hence  the  term, 
Principle,  or  "Method,  of  Least  Work."  (Footnote,  p.  30.) 

N.B.  If  the  extremities  of  bar  1,  supposed  to  be  in  tension 
in  above  proof,  were  displaced  in  directions  contrary  to  those 
assumed;  or  if  the  bar  were  in  compression,  no  matter  what 
the  actual  movement  of  the  extremities,  in  space,  due  to  the 
application  of  the  loading;  it  is  easily  shown  that  the  same 
result  would  be  reached  [eq.  (49)].  This  result  is  therefore 
general. 

29a.  Theorem  of  "  Least  Work  "  a  Particular  Case  of  a 
More  General  Relation.  The  use  of  eq.  (49)  has  already  been 
illustrated  in  solving  the  statically  indeterminate  frame  of 
§  28.  It  should  be  noted,  however,  that  the  idea  of  "  Least 
Work  "  which  has  given  rise  to  the  name  of  the  method  is 
purely  an  incidental  matter,  of  scientific,  rather  than  practical, 
interest;  since  the  proof  of  eq.  (49)  does  not  depend  upon  this 
idea,  but  is  based  on  the  ordinary  relations  of  geometry  and 
elasticity  in  connection  with  Castigliano's  Theorem  for  displace- 
ments of  the  joints  of  a  statically  determinate  structure.  This 
same  mode  of  proof  will  be  employed  later  (§  32)  in  establish- 
ing a  method  of  dealing  with  a  structure  containing  redundant 
bars  which  are  initially  too  long,  or  too  short,  to  fit  the  joints 
already  determined  by  the  "  necessary  "  bars  of  the  structure; 
and  it  will  be  seen  that  eq.  (49)  is  merely  a  particular  case  of 
this  more  general  relation. 

30.  Statically  Indeterminate  Structure  Containing  Two 
Redundant  Bars.  The  truss  in  Fig.  12,  resting  on  two  smooth 
horizontal  piers  at  the  same  level,  consists  of  one  square  panel 
and  one  rectangular  panel,  involving  eleven  bars  fitted  origi- 
nally (i.e.,  before  the  loading)  without  strain.  There  are  three 
equal  loads,  each  equal  to  P,  applied  at  the  three  upper  joints. 
Evidently  there  are  two  redundant  or  "  unnecessary  "  bars; 
that  is,  no  more  than  two  bars  could  be  omitted  without  collapse. 
For  example,  bars  5  and  9  could  be  omitted;  or  bars  5  and  2; 
and  the  remainder  of  the  structure  would  hold  its  shape  and 
form  a  "  statically  determinate  "  structure.  (In  this  particular 
case  it  also  happens  that,  since  bars  6  and  11  are  in  the  same 


34 


MECHANICS    OF    INTERNAL    WORK 


30 


line  and  there  is  no  load  at  their  junction,  the  three  bars  1,  2, 
and  7  might  be  omitted  without  collapse.) 

Let  us  imagine  bars  1  and  2  removed  and  the  (unknown) 
stresses  in  them  (suppose  each  to  be  a  tension),  TI  and  T2, 
to  act  at  the  proper  joints  on  the  remainder  of  the  truss,  this 


T 


FIG.  12. 

remainder  being  now  a  "  statically  determinate  "  structure  under 
the  nine  external  forces  (see  Fig.  13),  TI  in  two  places,  T2  in 
two  places,  P  in  three  places  (loads),  and  the  two  vertical 
reactions  fP  and  £P. 

We  can  now  by  simple  statics  determine  the  values  of  the 
internal  stresses  T3,  T4,  etc.,  up  to  T\\  (inclusive)  in  terms  of 
these  "  external  forces "  just  mentioned.  These  values  are 
assembled  in  the  following  table  and  also  (for  a  future  purpose) 


FIG.  13. 


values  of  the  derivatives  of  each  of  these  internal  stresses  with 
respect  to  T\  and  to  TV  We  here  denote  cos  a  by  m,  and 
sin  a  by  n  (a  itself  is  33°  40'). 


30 


TWO  REDUNDANT  BARS  IN  FRAME 

1 


35 


T10  = 


_±  •-IZp-wfc 


•  =  0 


=  0 


dT2 
dT7 


dT^ 
dT2 

dTio 


n 


dT 


n 


(50) 


Let  us  now  restore  bar  2  to  the  truss  in  Fig.  11.  The  result 
is  a  new  truss  with  one  redundant  bar  (taken  as  bar  2)  the 
stress  in  which  (T2)  can  therefore  be  found  in  terms  of  TI 
and  P  by  §  29;  since  for  this  new  truss  the  external  forces 
are  the  two  TI'S,  the  three  P's,  the  fP  and  the  £P;  (Ti  is  yet 
unknown,  of  course). 

We  now  put  — Vj  '  n  =0,  with  the  understanding  that 
dJ.  2 

U2  .  .  .  n,  denotes  the  total  internal  work  of  this  new  truss, 
consisting  of  bars  2  to  11  inclusive;  that  is, 


U 


2  • 


11  = 


. 

I-...H 


,,n 
.,  •  •  (51) 


whence 

dU2... 

dTo 


•  +OnTir%r9  =0.     (52) 

CH  2 


36  MECHANICS   OF    INTERNAL   WORK  §  30 

Filling  out  details  of  eq.  (52)  by  the  aid  of  the  table,  or 
eqs.  (50),  there  is  obtained,  after  reduction, 


-  +[C72  +n*C7 

]    =  0,  .     .«  ...   .  V    .     (53) 


which  contains  two  unknowns,  viz.,  TI  and  T2. 

Similarly,  restore  bar  1  to  the  truss  in  Fig.  13,  and  we  again 
have  a  new  truss  with  only  one  redundant  bar  (taken  as  bar  1). 
The  new  truss  consists  of  bar  1  and  bars  3  to  11  inclusive,  the 
forces  external  to  it  being  the  two  TYs,  the  three  P's  and  the 
fP  and  £P.  Its  total  internal  work,  which  may  be  denoted 
by  Ui,  3  .  .  .  11, 


2  2  2 

and  by  putting  its  first  derivative  with  respect  to  T\  equal  to 
zero  we  obtain  [by  eq.  (49)]  an  expression  for  T\  in  terms  of 
the  external  forces  just  mentioned.  Now 

dUi>*--"_CTdT*  lC  T  ^J 

*dT.4.  ,   dTi 


...     (55) 

and  this  is  to  be  placed  equal  to  zero.     Therefore,  substituting 
details  from  eqs.  (50),  we  have,  after  simplification, 


0.  .     (56) 

We  now  have  two  equations,  viz.,  (53)  and  (56),  from  which  by 
ordinary  algebra  both  TI  and  T2  can  be  determined. 

If  we  take,  as  a  particular  instance  of  this  example,  the 
value  of  45°  for  a,  we  must  put  \/2  -=-2  both  for  m  and  n,  while 
JP  replaces  the  fP  and  also  the  fP  in  eqs.  (50),  TI  will  also 
be  equal  to  T%,  from  symmetry;  and  .*,  eq.  (56)  alone  will 


§  32         MORE  THAN  TWO  REDUNDANT  BARS  37 

serve  to  determine  T\,  which  in  this  way  is  found  to  be  (with 
all  the  O's  taken  equal) 

(57) 


31.  Statically  Indeterminate  Structure  with  More  than 
Two  Redundant  Bars.  (The  usual  stipulation  is  made,  that 
all  the  bars  are  closely  fitted  originally,  without  strain.)  With 
three  redundant  bars,  say,  bars  1,  2,  and  3,  we  have  simply 
to  replace  them  by  the  pairs  of  (unknown)  forces  which  they 
exert  on  the  joints  at  their  extremities,  and  by  applying  ordi- 
nary statics  obtain  the  stresses  T±,  T&,  etc.,  in  all  the  other 
bars  in  terms  of  the  unknown  T\,  T2,  and  T%,  and  the  external 
forces  of  the  original  truss.  The  derivatives  of  each  of  the 
stresses  T±,  T&,  etc.,  are  then  obtained  with  respect  to  T\,  to 
T2,  and  to  !T3,  and  set  down  in  a  table. 

Bar  1  is  then  replaced  in  the  truss,  which  thus  becomes  a 
truss  of  one  redundant  member  (viz.,  bar  1)  its  external  forces 
consisting  of  the  two  TVs,  the  two  TS'S,  and  the  external  forces 
of  the  original  truss.  By  the  use  of  eq.  (49)  an  expression  is 
obtained  for  T\  in  terms  of  T2,  T3,  and  the  original  external 
forces. 

By  replacing  bar  2  in  the  truss  instead  of  bar  1  and  using 
eq.  (49),  a  second  equation  is  obtained  giving  T^  in  terms  of 
Ti  and  jP3,  and  the  original  external  forces.  Similar  treatment 
with  bar  3  gives  a  third  equation  furnishing  Ts  in  terms  of 
T\,  T%,  and  the  original  external  forces.  From  these  three 
independent  equations  the  values  of  the  three  quantities  T\, 
TI,  and  TS  are  then  easily  determined. 

The  procedure  is  now  evident  for  the  case  of  .any  number 
of  redundant  bars,  but  manifestly  involves  a  large  amount  of 
detail  when  there  are  more  than  two  such  bars. 

32.  Statically  Indeterminate  Structure  with  One  Redun- 
dant Bar,  this  Bar  being  Originally  Too  Short,  or  Too  Long, 
to  Fit  into  its  Place.  In  a  structure  of  one  redundant  bar 
let  us  suppose  that  the  "  necessary  "  bars  have  first  been  fitted 
together  and  that  is  it  then  found  that  the  remaining  (redundant) 
bar  is  ^  inches  longer,  or  shorter,  than  the  distance  between 
the  joints  which  it  is  to  connect;  so  that  to  adjust  it  to  its 


38 


MECHANICS   OF    INTERNAL   WORK 


§32 


place  it  is  necessary  by  external  means  to  force  the  two  joints 
nearer  to,  or  further  from,  each  other,  and  either  to  stretch  or 
shorten  the  bar  itself,  to  cause  its  extremities  to  fit  over  the 
pins  at  these  joints.  This  being  done,  and  the  external  con- 
straint removed,  the  structure  is  provided  with  some  character 
of  support,  and  is  loaded  at  one  or  more  joints.  The  small 
distance  ^  (surplus  or  deficiency,  in  length;  of  the  bar  in 
question)  is  supposed  to  be  small;  so  that  when  the  bar  is 
forced  into  position  and  the  loads  placed,  the  elastic  limit  is 
not  passed  in  any  bar.  We  are  now  to  prove,  denoting  by  U 
the  total  internal  work  of  the  structure  (all  the  bars)  in  its 
final  condition  under  load,  and  by  TI  the  stress  in  the  redundant 
bar  (bar  1),  that 

-*,,    ...,..  .  (57) 


for  the  case  where  bar  1  is  assumed  to  be  in  tension  and  is 
originally  too  short. 

A  specific  case  will  be  taken.     In  Fig.  14  the  five  "  neces- 
sary "  bars  (2,  3,  4,  5,  and  6)  when  fitted  together  form  the 


d" 


FIG.  14. 

quadrilateral  figure  (dotted  lines)  abed]  bar  4  being  a  diagonal, 
bd.  Bar  1,  whose  unstrained  length  is  oc',  is  to  form  the  other 
diagonal;  that  is,  it  is  _tp_  connect  the  joints  a  and  c;  but  is 
too  short,  by  an  amount  c'c,  or  ^  inches.  Joint  b  is  fixed  on  the 
support  A,  while  the  pin  of  joint  d  is  free  to  slide  along  the 


§  32  REDUNDANT  BAR  TOO  SHORT  39 

smooth  hard  surface  of  the  fixed  support  /).  Let  now  the 
lower  extremity  of  bar  1  (oc')  be  forcibly  fitted  over  the  pin 
of  joint  c  and  the  load  P2  applied  at  the  latter  joint.  This 
special  external  constraint  having  been  removed,  and  the  load 
P2  being  in  place,  we  find  joint  d  in  a  new  position,  d"  ',  on 
the  supporting  surface  of  Z);  while  joints  a  and  c  are  now  found, 
respectively,  at  a"  and  c"  '.  Bar  1  is  in  a  state  of  tension,  the 
stress  in  it  being  Ti,  unknown.  Its  unstrained  length  being 
ac'  =  li,  its  present  length  is  a"c"  =  Z  +  ^i,  where  AI  is  its  elonga- 
tion. 

Let  us  now  consider  bar  1  to  be  removed,  its  action  on  the 
two  joints  a"  and  c"  being  fully  represented  by  the  forces  T 
at  a  and  T'  at  c",  these  forces  being  in  the  same  line  a"c", 
each  equal  to  the  unknown  TI,  and  pointing  toward  each  other, 
as  shown.  They  may  now  be  considered  as  external  forces 
(and  independent  of  each  other),  acting  on  the  remaining  bars, 
viz.,  2,  3,  4,  5,  and  6;  these  bars  forming  a  statically  deter- 
minate structure  which  has  been  changed  from  the  form  abed 
(dotted  lines)  into  the  form  a"bc"d"  (full  lines)  by  the  action 
of  the  external  forces  T,  T',  and  P2;  the  supports  being  of 
the  nature  described. 

Hence,  if  U"  denote  the  internal  work  for  all  the  bars  except 
bar  1,  we  have,  from  Castigliano's  Theorem  (§  21)  the  dis- 
placement of  a  in  the  direction  of  force  T7,  viz.,  the  projection 
marked  y  in  the  figure  is 


and  similarly,  the  displacement  of  c  in  direction  of  force  T', 
or  the  projection  y'  of  the  figure,  is 


(Note  that  for  this  figure  both  of  the  above  signs  are  +r 
since  each  of  the  displacements  concerned  takes  place  in  the 
direction  in  which  the  corresponding  force  points.) 

But,  from  the  geometry  of  the  figure,  oc'  +  ^  =  2/+a"c"  +  ?/;. 

Mi+V'.  ((>°) 


40  MECHANICS    OF    INTERNAL    WORK  §  32 

/  T 

Introducing  into  (60)  the  relation  *i—  gr^K  =Ci7V  and 

"  1^1 

the  values  of  y  and  yf  from  (58)  and  (59),  we  have 

7!^.    .    .'•".    ....     (61) 


Now  T  and  T'  each  =T\  and  CiTi  is  nothing  more  than 
the  derivative  of  the  internal  work  of  bar  1,  viz.,  JCiTi2, 
with  respect  to  T\.  Hence  the  left-hand  member  of  eq.  (61)  is 
the  complete  derivative  of  the  internal  work,  U,  of  all  the  bars 
(including  bar  1)  and  (61)  may  therefore  be  written 

;#r=4>-       ....    j-  •    -     (62) 

The  value  of  the  stress  TI  may  therefore  be  determined  by 
filling  out  eq.  (62)  and  solving  for  T\',  i.e.,  by  first  expressing 
the  stress  in  each  bar  in  terms  of  T\  (use  being  made  of  the 
"  free  body  "  (full  Hnes)  in  Fig.  14;  the  values  T  and  T'  being 
now  replaced  by  jPi);  and  then  writing  out  the  separate  deriv- 
atives needed  (besides  the  stresses  thesmelves)  for  substitution 
in  the  six  terms  in  the  left-hand  member  of  eq.  (62),  i.e.,  in 
the  expression 


=*0.   .    (63) 

While  it  is  evident  that  in  case  bar  1  is  originally  too  short 
to  fit  into  its  place  it  will  be  found  in  a  state  of  tension  after 
it  has  been  forcibly  fitted  into  place  and  the  truss  left  to  itself 
without  load;  nevertheless,  its  place  in  the  truss  and  the 
position  of  the  load  or  loads  may  be  such  that  afterwards,  when 
the  truss  is  loaded,  it  will  be  found  in  a  state  of  compression, 
instead  of  tension.  This,  result  would  be  evidenced  by  the 
obtaining  of  a  negative  result  for  the  stress  TI  in  the  final 
solution  of  eq.  (63);  that  is,  TI  is  thus  found  to  be  a  "  negative 
tension,"  or  compressive  stress. 

Bar  1  Originally  too  Long.  If  the  redundant  bar  (bar  1) 
is  originally  too  long,  by  ^  inches,  to  connect  joints  a  and  c 
of  Fig.  14,  point  c'  will  be  below  c"  .  We  have,  therefore,  with 
bar  1  in  its  final  position  and  the  truss  loaded,  the  bar  being 


§  33  REDUNDANT  BAR  TOO  SHORT  41 

•assumed  in  a  state  of  compression,  and  its  final  shortening  being 

^ClTl, 

ac,  =l-*o,=y  +  (l-h)+y'.    .    •    ,    .     (64) 

If  we  now  suppose  bar  1  to  be  removed,  the  forces  T  and  Tf 
(at  a"  and  c")  taking  its  place  for  equilibrium,  these  two  forces 
must  be  inserted  pointing  away  from,  instead  of  toward,  each 
other;  so  that  we  now  have 

dV"                        dU" 
y  =  —ftf    and    t/'  =  —^ (65) 

Hence,  by  combining,  we  reach  the  same  result  as  in  the  case 
of  the  bar  being  too  short,  viz., 

dU"    dU" 

~~^~^^~      r^1==^°7 


or 

^r=*>;      .  (67) 

but  it  must  be  noted,  that  in  the  use  of  this  relation  for  the 
case  of  bar  1  too  long,  the  stress  in  the  bar  must  be  assumed  to  be 
compressive  and  the  determination  of  stresses  in  the  other 
bars  in  terms  of  T\  must  be  made  on  this  basis.  (If  the  actual 
stress  in  bar  1  is  tensile  the  fact  will  be  brought  out  by  the 
obtaining  of  a  negative  value  for  T\  in  the  final  numerical 
result.) 

It  will  be  noted  that  when  ^  =  zero  for  the  redundant  bar 
eqs.  (62),  (63),  and  (67)  reduce  to  what  has  been  called  the 
Theorem  of  "  Least  Work  "  (see  §  29a). 

33.  Example  of  Single  Redundant  Bar,  when  Originally 
Too  Short.  For  this  we  may  take  the  frame  of  six  bars  already 
shown  in  Fig.  9  (p.  26),  §  27,  and  suppose  that  originally 
bar  1  was  too  short,  by  ^  inches,  to  connect  joints  a  and  e. 
In  the  treatment  of  the  problem  in  §  27  the  stresses  in  the  other 
bars  were  found  in  terms  of  T\  (there  called  T  and  T',  tem- 
porarily) on  the  assumption  that  the  final  state  of  bar  1  is 
tension]  which  is  suitable  for  present  purposes  in  the  use  of 
eq.  (57),  or  (63),  according  to  which  we  have  only  to  set  the 


42  MECHANICS    OF    INTERNAL    WORK  §  34 

expression  in  eq.  (44)  equal  to  fa,  instead  of  zero.     This  being 
done,  the  result  is 


~ 


which  may  now  be  compared  with  the  value  in  eq.  (42)  for  the 
case  where  bar  1  could  be  fitted  in  place  originally  without 
strain. 

34.  Statically  Indeterminate  Structure  with  Two  or  More 
Redundant  Bars  ;  Each  of  which  is  Too  Long,  or  Too  Short, 
to  Fit  into  Place.  A  case  of  two  such  bars  will  suffice  to 
indicate  the  method  for  any  other  number.  Let  the  structure 
contain  nine  bars,  of  which  only  seven  are  "  necessary  "  bars, 
the  other  two  being  redundant.  Let  the  bars  to  be  taken  as  the 
the  redundant  bars  be  4  and  5;  bar  4  being  originally  (^0)4 
inches  too  short,  and  bar  5  (fa)  5  inches  too  long,  to  fit  into 
place.  Assume,  therefore,  that  the  final  state  of  4  is  tension; 
and  that  of  5,  compression. 

Let  now  the  redundant  bars  4  and  5  be  removed  from  the 
truss,  and  the  forces  which  they  exerted  on  the  joints  at  their 
extremities  put  in;  T±  at  two  joints,  and  T$  at  two  joints, 
pointing  in  proper  directions  to  correspond  with  the  assumption 
of  tension  or  compression  in  the  corresponding  bars.  We  now 
have  a  statically  determinate  structure  consisting  of  the  seven 
bars  1,  2,  3,  6,  7,  8,  and  9;  the  stresses  in  which  are  now 
obtained  in  terms  of  the  loads  and  of  the  unknown  T±  and  T$r 
by  ordinary  statics.  Now  conceive  bar  4  to  be  again  in  place 
(but  not  bar  5),  and  we  have  a  statically  indeterminate  struc- 
ture, with  only  one  redundant  bar  (bar  4),  the  external  forces 
for  which  are  the  two  TYs,  the  loads,  and  the  reactions  of 
supports;  in  terms  of  which  forces,  T±  may  now  be  obtained  by 
the  use  of  eq.  (62),  viz.  : 

d[Ui,  2,  3,  4;    6,  7,  8,  Q]        ,  .  v  ,aQ. 

-  •  • 


where    the    left    member    consists    of    terms    of    the    type 

dT 

-m-,  one  for  each  of  the  original  bars,  except  5. 


§  35     REDUNDANT  BARS  TOO  SHORT  OR  TOO  LONG       43 

Again,  let  bar  5  be  put  back  in  the  truss,  and  bar  4  taken 
out;  and  the  structure  thus  obtained  contains  only  one  redun- 
(  lant  bar  (which  now  is  bar  5)  and  is  acted  on  by  external  forces, 
consisting  of  the  two  TYs,  the  loads,  and  the  reactions  of 
supports.  Hence  from  eq.  (67), 

d[Ul,  2,  3;    5,   6,   7,  8,  Q]        ,  .   . 

.....     (59) 


where    the    left-hand    member    consists    of    terms    of    the 

dT 
type  CT-T^r,  for  one  each  of  the  bars  of  the  original  truss, 

except  4.  Eq.  (69)  will  give  the  stress  T5  in  terms  of  the 
external  forces  just  mentioned. 

Equations  (68)  and  (69),  then,  are  two  simultaneous  equa- 
tions containing  the  two  unknown  quantities  T  '4  and  T^  ;  and 
the  values  of  the  latter  are  now  obtained  by  ordinary  elimina- 
tion. Should  a  negative  number  be  obtained  as  the  value  of 
cither  T±  or  T$  it  indicates  that  the  character  of  the  stress  in 
the  bar  is  the  opposite  of  that  originally  assumed. 

The  procedure  to  be  adopted  in  case  there  are  more  than 
two  redundant  bars  is  now  evident.  There  will  be  as  many 
independent  equations  like  (68)  or  (69)  as  there  are  redundant 
bars.  As  a  basis  for  each  one  of  these  equations  we  have  under 
consideration  a  truss  with  only  one  redundant  bar,  the  forces 
external  to  this  truss  being  the  original  loads,  the  reactions  of 
supports,  and  the  forces  exerted  (on  the  joints  at  their  extrem- 
ities) by  all  the  other  (original)  redundant  bars.  Attention  is 
again  called  to  the  necessity  of  assuming  each  of  the  original 
redundant  bars  to  be  in  tension,  or  in  compression,  according 
to  whether  it  is  too  short,  or  too  long  (respectively)  to  fit  into 
place  after  the  "  necessary  '"  bars  have  been  put  together. 

35.  Temperature  Stresses  in  a  System  of  Elastic  Bars. 
The  foregoing  treatment  finds  application  in  the  following 
circumstances:  A  change  of  temperature  from  that  common  to 
all  the  bars  at  the  time  of  assembling  the  truss,  if  we  suppose 
an  original  close  fit,  without  strain,  at  that  temperature,  may 
create  stresses  in  the  bars  without  any  load  being  placed  on  the 
truss. 


44  MECHANICS   OF    INTERNAL    WORK  §  35 

There  may  be  redundant  bars,  or  the  character  of  the 
supports  may  be  such,  that  when  the  temperature  rises  (or 
falls),  the  lengths  natural  to  the  bars  at  the  new  temperature 
(with  no  load  on  truss)  are  not  those  that  they  are  constrained 
to  have  in  their  connection  with  each  other;  so  that  if  the 
truss  were  reassembled  at  the  new  temperature  the  bars  would 
not  fit  into  place  without  special  external  (temporary)  constraint ; 
that  is,  after  the  "  necessary  "  bars  had  been  put  together  at 
the  new  temperature  each  of  the  others  would  be  found  to  be 
too  short,  or  too  long,  to  fit  into  place  without  such  external 
constraint. 

Hence  we  may  determine  the  stresses  in  these  latter  (redun- 
dant) bars  (and  ultimately  in  all  the  bars),  after  noting  by 
how  much  each  such  bar  is  too  short,  or  too  long,  to  fit  into  its 
place  in  the  assemblage  of  the  "  necessary  "  bars,  by  applying 
the  methods  of  §§32,  33,  and  34.  Stresses  thus  induced  are 
called  temperature  stresses.  These  stresses  are  usually  obtained 
independently  of  the  action  of  loads,  and  combined  later  with 
those  due  to  the  loading. 


§  36        FLEXUKE  OF  BEAMS.   INTERNAL  WORK         45 

CHAPTER  IV 

INTERNAL  WORK  IN  BEAMS  UNDER  FLEXURE.    APPLICATIONS 

36.  Internal  Work  of  a  Bent  Beam.  Straight  Beams; 
and  Curved  Beams  of  Large  Radius  of  Curvature.  When  a 
beam  originally  straight,  or  curved  (and  in  that  case  the  radial 
thickness  must  be  small  compared  with  the  radius  of  curvature), 
is  slightly  deformed  by  being  made  the  means  of  holding  in 
equilibrium  a  system  of  applied  forces  and  reactions  (all  acting 
in  the  same  plane,  containing  the  axis  of  the  beam),  each  slice 
or  lamina  (of  the  beam),  bounded  by  two  imaginary  cu'tting 
planes  at  right  angles  to  the  axis  of  the  beam  and  a  certain 
small  distance  apart  denoted  by  ds,  measured  along  that  axis, 
consists  of  a  bundle  of  fibers  (so  called)  on  the  extremities  of 
which  (in  the  two  cutting  planes),  by  the  common  theory  of 
flexure,  we  find  elastic  forces  acting,  viz.,  tensile,  or  com- 
pressive,  and  shearing.  In  general  [see  Fig.  15,  (A)],  we  find 
at  each  end  of  the  lamina,  or  block,  a  total  shear  represented 
by  J  (Ibs.)  (or  «/'),  a  total  thrust,  T  (Ibs.)  (or  T'),  consisting  of 
a  set  of  uniformly  distributed  compressive  or  tensile  forces  of 
uniform  intensity  p\  (Ibs.  per  sq.in.) ;  and  a  set  of  tensions  and 
compressions  forming  a  "  stress-couple  "  whose  moment  ("  bend- 

P2^ 

ing  moment  ")  is  denoted  by  M  and  is  equal  to  —  (in  which  p2 

t/ 

denotes  the  stress,  Ibs.  per  sq.in.,  in  the  outer  fiber  at  a  distance 
e  from  the  center  of  gravity,  or  axis,  of  the  cross-section,  and  / 
the  "  moment  of  inertia  "  of  the  plane  figure  (formed  by  the 
cross-section)  with  reference  to  the  gravity  axis  G,  perpendicular 
to  the  paper.  dF  denotes  an  elementary  area  of  the  cross- 
section  and  z  is  its  distance  from  the  gravity  axis  G.  (See 
p.  354  of  the  author's  "  Mechanics.") 

Before  stress,  the  block  had  the  unstrained  shape  NHHQNo. 
Under  stress,  however,  its  shape  is  NnH'fHQN^  which  now  shape 
we  may  consider  to  have  come  about  thus:  The  uniformly 
distributed  thrust  T  causes  all  the  fibers  to  shorten  equally  by 


46 


MECHANICS    OF    INTERNAL    WORK 


36 


an  amount  cUi,  thus  bringing  the  face  NH  to  a  parallel  position 
N'H';  then  the  couple  acts,  causing  the  fibers  to  shorten  or 
lengthen  unequally,  but  proportionally  to  the  distance  z  from 
the  gravity  axis  G  of  the  cross-section  (in  which  operation  the 
outer  fiber  NN-0  shortens  by  an  amount  dX2) ;  then  the  shearing 
stress  comes  into  play  to  distort  slightly  the  corner  angles  of 
the  small  blocks  into  which  each  fiber  may  be  conceived  to  be 


T' 


J\  ,1  ^ 

VKsJ*" 


(B) 


subdivided  (no  attempt  has  been  made  to  show  this  distortion 
in  the  figure). 

We  have  therefore, 

The  work  of  thrust  on  «d,  or  one  fiber,  =  \p\dF  .  ab  =  [%p\dF}dX\  ; 

but,  from  definition,  -~  =  Q  ; 

hence  the  work  of  thrust  (one  fiber),    =  ^p\dF-  =  (? 


hence  the  whole  work  of  thrust  for  all  the  fibers  composing  the 
lamina  is  (with  F  denoting  whole  area  of  section  N"H"), 


(70) 


As  to  the  work  of  the  stress-couple,  each  fiber  shortens  an 
of  d^  due  to  a  force  —p-2dF',    and  hence 

fiber  =(%.-pd,F\(-dx2\.    But    ^r»|r, 

\  w       />  •*  /   \  p  I  r\  Q         H. 

\  O  /     \t/  /  (*o  /^ 


amount  be,  =—  of  dfa,  due  to  a  force  —p-2dF]    and  hence   the 

6  & 


work    on    one 


and 


§  36  FLEXURE.      INTERNAL    WORK  47 

therefore  the  work  on  one    fiber,  as  due  to  the  stress-couple,  is 

p22ds    dFz2  f 

J  .      2   •  .  —p—  ;   and  therefore  the 

Whole  work  due  to  stress-couple  1  _  1    P^ds  CN"  ,  ™  2  _  ds    p# 
for  the  whole  lamina  or  block   J  =  *  '  ~~E#~  )H,,  d        =  ~2  '  Ee*  ' 


which  can  be  written  pr^rf  {  -  )  .     But  -  is  M.  the  moment  of 
ZHjl  \  e  /  e 

the  stress-couple  (or  the  "  bending  moment  ")  and  hence  the  last 
expression  becomes 


As  for  the  work  of  the  shear,  J,  if  the  whole  block  were 
gradually  distorted  from  a  square-cornered  condition  to  a 
rhombus  [d  being  the  change  of  angle  at  the  corner.  See 
Fig.  15  (B)]  by  the  shearing  couple  whose  moment  is  Jds  we 
should  have  the  work  of  J  (all  the  other  forces  being  neutral) 

=  J  .  Jab=*y(ds)d,  where  d  is  the  angle  of  shearing  distortion. 

Now   (by  definition)   d  =  -p~E8,  where  Es  is  the  modulus  of 
elasticity  for  shearing  (p.  227,  M.  of  E.),  and  therefore  the  whole 

J2 

work  due  to  shear  for  this  lamina  or  block  would  be  i  .  TTTT  .  ds. 


But  since  the  shear  is  not  uniformly  distributed  over  the  cross- 
section  in  any  case  of  flexure,  this  last  expression  must  be 
modified  by  the  use  of  a  coefficient,  or  multiplier,  A,  to  suit 
different  forms  of  cross-section.  For  example,  for  a  solid 
rectangular  cross-section  with  two  bases  parallel  to  the  "  force 
plane  "  or  plane  of  the  paper,  it  can  be  proved  that  A=f  ;  while 
for  a  solid  circular  section  A=^-.  For  an  ordinary  I-beam 
with  a  thin  web,  if  the  web  be  considered  as  bearing  most  of 
the  shear,  A  becomes  practically  unity  (but  then  F  is  the  area 
of  the  section  of  the  web  alone).  Hence,  the  work  of  the  shear 
for  one  lamina,  of  length  ds,  would  be 

-   ^  (72) 

2  -        ........    (72) 


48  MECHANICS    OF    INTERNAL   WORK  §  37 

The  forces  shown  in  Fig.  15  (A),  as  acting  on  the  slice  or 
lamina  situated  between  two  neighboring  cross-sections,  play 
the  part  of  external  forces  for  this  elastic  body.  But  since  this 
elastic  body  may  be  considered  to  be  made  up  of  an  immense 
number  of  elastic  bars  pivoted  to  each  other,  these  bars  being 
gradually  stretched  or  shortened  during  the  gradual  application 
of  external  forces,  the  total  internal  work  performed  in  this 
operation  will  be  equal  in  value  to  the  sum  of  the  three  terms 
just  found,  viz.  : 

T2ds       M2ds     A  J2ds 


Now  an  entire  beam,  straight  or  curved,  may  be  considered 
to  be  made  up  of  a  great  number  of  consecutive  slices  or 
laminae  like  that  in  Fig.  15  (A),  each  having  a  length  ds  along 
the  axis  of  the  beam  and  subjected  to  a  thrust  T  at  each  end; 
to  two  stress-couples,  one  at  each  end,  of  a  moment  M;  and  to 
two  shears,  one  at  each  end.  Consequently,  the  total  internal 
work  of  such  a  beam  can  be  expressed  as 

T*ds         M2ds      rAJ2ds 


(the  integration  being  extended  along  the  whole  length  of  the 
beam)  and  may  be  placed  equal  to  the  sum  of  the  amounts  of 
external  work  of  the  external  forces  applied  to  the  beam,  or 
"  applied  forces  "  (including  reactions)  to  which  the  state  of 
stress  of  the  beam  is  due. 

37.  Displacement  of  a  Point  in  Axis  of  a  Straight  Prismatic 
Beam  in  Flexure.  The  expression  for  U  in  eq.  (74),  having  been 
once  found  in  any  given  case,  may  be  used  inCastigliano's  Theorem 
[§21,eq.  (19)]  for  determining  deflections  or  displacements  of  any 
point  of  the  axis  of  a  beam  where  there  is  a  load  applied  (or 
where  a  load  may  be  imagined  to  be  applied  and  then  reduced 
to  zero-,  as  indicated  in  §  23a).  The  term  due  to  shear  is 
generally  of  very  small  consequence  and  may  therefore  be 
omitted  in  many  cases  without  sensible  error.  The  term 
involving  the  thrust  is  also  sometimes  negligible,  according  to 
circumstances.  The  beams  concerned  being  usually  of  homo- 
geneous material,  the  quantity  E  may  then  be  taken  outside 


§  38  FLEXURE.   INTERNAL  WORK  49 

the  integral  sign;  and,  furthermore,  if  all  cross-sections  are 
alike  and  similarly  placed,  the  "  moment  of  inertia,"  /,  of  the 
cross-section  will  be  the  same  at  all  sections  and  consequently 
may  in  such  cases  be  placed  outside  the  sign  of  integration. 
Frequently  when  it  is  wished  to  differentiate  this  expression 
for  U  with  reference  to  some  quantity  which  it  will  eventually 
contain  (say,  one  of  the  external  forces  P),  the  operation  may 
be  shortened  by  differentiating  the  general  expression  under  the 
integral  sign;  thus  (in  the  term  involving  M) 

d   riM2ds         l  2M   dM\   ds 


Hence,  when  (later)  M  is  expressed  as  a  function  of  P  it  will 
simply  be  necessary  to  find  M,  and  (-jpO>  m  terms  of  P  and 

substitute  in  eq.  (75)  ;  thus  dispensing  with  the  squaring  of  the 
M  as  called  for  literally  in  eq.  (74).  A  similar  remark  applies 
to  the  other  general  terms,  involving  the  thrust  and  the  shear, 
respectively. 

38.  Example  I.  Horizontal,  Straight,  Prismatic  Beam. 
End  Supports.  Load  in  Middle.  Central  Deflection  Re- 
quired. The  (originally)  straight,  prismatic,  homogeneous  beam 
is  placed  on  two  supports  at  the  same  level  and  loaded  in  the 
middle  with  a  concentrated  load  P,  vertical.  Evidently  the 
reaction  'at  each  pier  is  V  =  P  +  2  (Fig.  16)  and  is  vertical 


1-f 

1" 

dx           i                                   1  c 

1 

\      !                                                                   1° 

M-  t  a.____ 

? 

1 

L  ^  

IE 

[ 

i 

|                                         7 

FIG.  16. 

(smooth  horizontal  surfaces);  and  the  thrust  is  zero  at  all  sec- 
tions; while  the  work  of  the  shear  will  be  neglected,  as  being 
very  small. 

It  is  required  to  find  the  vertical  deflection  y\  for  the  middle 
point  B  of  the  axis  of  the  beam,  as  due  to  the  gradual  applica- 
tion of  the  one  load  P,  applied  at  B.  Neglect  the  weight  of 


50  MECHANICS   OF   INTERNAL   WORK  §  38 

the  beam  itself.  Evidently  the  deflection  desired  is  the  dis- 
placement of  the  point  of  application  of  P  in  the  direction  of 
P  itself  and  therefore  can  be  obtained  by  an  application  of 
eq.  (25)  (i.e.,  Castigliano's  Theorem). 

In  this  case  we  shall  measure  the  abscissa,  x,  of  any  dx 
in  the  left-hand  half  of  the  beam,  from  the  extremity  0;  and 
since  the  beam  is  straight,  dx  will  take  the  place  of  ds  in  the 
general  eq.  (74).  The  "  free  body  "  of  length  x  is  acted  on 
by  the  reaction  JP  at  one  end  and  by  the  shear  J  and  the 
tensions  and  compressions  forming  the  stress-couple  of  moment 
M  at  the  other.  From  this  free  body,  taking  moments  about 
the  right-hand  extremity,  we  have  M  =  JPz,  and  hence  the 
internal  work  for  all  the  vertical  laminae  of  the  left-hand  half 
OB  of  the  beam : 


(76) 


SE1  \  3 

From  symmetry,  we  may  double  this  for  the  internal  work 
of  the  whole  beam  as  due  to  flexure  (that  due  to  shear  being 
neglected  and  that  due  to  thrust  being  zero  on  account  of  the 
zero  value  of  the  thrust  at  all  sections).  That  is,  the  total 
internal  work  for  the  beam  is 

E>273 

(77) 


Now  the  deflection  of  point  B  below  its  original  position  in  the 
horizontal  line  drawn  through  0  and  C,  is  the  displacement  of 
the  point  B  in  the  direction  of  the  external  force  P,  and  hence, 
by  eq.  (25), 

dU      .  d 


P  being  now  regarded  as  a  variable.    That  is, 
/   Z3    \  P/3 

*"=W)2P;  or  *-j&Bi>    •  •  •  (78) 

as  we  know  by  another  method.     (See  page  254,  Mech.  of  Eng.) 


§  39  FLEXURE.      INTERNAL   WORK  51 

39.  Influence  of  Shear  in  Preceding,  if  the  Beam  is  a 
Steel  I-Beam.  If  the  beam  in  foregoing  problem  in  a  steel 
I-beam  with  narrow  web  the  effect  of  shear  in  producing  deflec- 
tion may  not  be  negligible.  Considering  the  internal  work  of 
shear  as  well  as  that  of  the  stress-couples,  we  have,  since  the 
total  vertical  shear  in  this  case  is  /  =  JP  at  all  sections,  for 
the  internal  work  [see  eqs.  (74)  and  (77)], 


„ 


A(P\* 
2V27   ' 


in  which  Fw  denotes  the  sectional  area  of  the  web  alone  while 
the  coefficient  A  may  be  taken  as  unity.     Therefore 

dU      PI3         PI 


As  a  numerical  instance,  let  us  take  the  case  of  a  10-inch 
Cambria  steel  I-beam,  weighing  25  Ibs.  per  foot,  and  of  a  length 
of  10  feet;  to  be  supported  as  in  Fig.  16  and  loaded  at  middle 
with  P  =  10,000  Ibs.;  (the  web  placed  in  the  usual  vertical 
position). 

For  the  cross-section  of  this  beam  we  find  from  the  Cambria 
Co.'s  hand-book  that  I  (about  neutral  axis  perpendicular  to 
the  web)  =122.1  in.4  while  ^  =  7.85X0.31  =  2.43  in.2  Young's 
modulus  may  be  taken  as  £7  =  30,000,000  lbs./in.2;  and  E8,  the 
modulus  of  elasticity  for  shearing,  as  E8  =  l  1,500,000  lbs./in.2 
With  these  values  eq.  (80)  gives 

t/i  =  0.0982  +  0.0107  =  0.  1089  inch. 

The  portion  (viz.,  0.0107)  due  to  shearing  action  is  seen  to 
be  about  ten  per  cent  of  the  whole.*  (The  weight  of  the  beam 
itself  has  been  neglected.)  The  effect  of  the  shear  would  be 
much  less  marked  with  a  beam  of  rectangular  section;  and  also 
in  the  present  case  of  the  I-beam  if  the  span  were  considerably 
increased. 

*  An  interesting  case  of  this  nature  is  worked  out  by  Professor  C.  J. 
Tilden  in  the  Engineering  News  of  Feb.  24,  1910,  p.  228. 


52 


MECHANICS   OF   INTERNAL   WORK 


40 


40.  Example  II.  Deflection  of  Straight  Prismatic  Beam, 
Ends  Supported,  Load  Uniform.  (Fig.  17.)  An  (originally) 
straight  homogeneous,  prismatic  beam  rests  on  two  supports 


UlllllllllilJ! 


Si 


FIG.  17. 

at  its  extremities  at  the  same  level,  bearing  a  uniformly  dis- 
tributed load  over  its  whole  length  of  an  amount  W  =  wl  (w  — 
load  per  linear  unit;  I  being  the  whole  length).  It  is  required 
to  find  the  vertical  deflection  of  the  middle  point  B.  Since 
there  is  no  finite  vertical  load  at  this  point  let  us  imagine  one 
(=P,  say)  to  be  placed  there  and  finally  reduce  it  to  zero.  We 
then  have  the  reaction  at  each  support  V  =  $(W+P).  Now 
for  any  vertical  lamina  of  thickness  dx  of  the  beam  the  moment 
of  the  stress-couple  is 

--*-,  (81) 


x  being  measured  from  the  left-hand  support  0.  There  is  no 
thrust.  Neglect  the  work  of  shear.  To  secure  greater  brevity 
let  .us  use  the  form 


fo    fdM\ds 
Jo  M(dp)Tl> 


mstead  of 


d  f  CcM*ds 


for  the  deflection  of  the  point  of  application  of  the  load  P. 
Since  the  summation  having  to  do  with  the  dx's  in  the  right- 
hand  half  of  the  beam  is,  from  symmetry,  equal  to  that  in  the 
left-hand  half,  we  may  double  the  expression  applying  to  the 
left-hand  half.  Hence,  noting  that  in  this  case  ds  =  dx,  and 


that,  for  any  dx  on  OB,  M  = 
have  [see  eq.  (75)]: 


wx 


dM    x 


— -;  and        =    ;    we 


§  41  DEFLECTION    OF  BEAMS  53 

2 


, 

dpdx 


---+   •  (82) 


This  gives  the  deflection  of  B  as  due  to  the  distributed  load 
W  and  the  concentrated  load  P  at  the  middle.  We  have  only 
to  make  P  equal  to  zero  to  obtain  the  deflection  B  for  the  actual 

5      Wl3 
case  of  the  distributed  load  alone,  that  is,  2/i  =  ooj  •  ~gT- 

Note.    The  "  dummy  "  force  P  having  served  its  purpose 

dM      x 
in  enabling  us  to  get  -jp-,  =^,  we  might  have  put  P  equal  to 

zero  in  the  expression  for  M  above,  before  substituting  in  the 
integration,  and  it  is  also  interesting  to  note  that  the  value 

of  -Tp-  is  the  same  as  the  moment  at  any  section  due  to  a  load 

of  unity  applied  vertically  at  the  point  B  (thus  leading  to 
"  Fraenkel's  Formula."  *  Of  course,  also,  by  putting  W=  zero 
in  eq.  (82)  we  may  obtain  the  deflection  due  to  P  alone  and 
thus  check  the  result  already  obtained  in  eq.  (78). 

41.  Example  III.  Deflection.  Beam  with  Overhang.  A 
continuous  prismatic  beam  (homogeneous  and  originally 
straight)  rests  on  two  unyielding  supports  at  the  same  level 
as  shown  in  Fig.  18.  The  vertical  force  P  is  applied  at  D, 
midway  between  0  and  B,  while  the  extremity  of  the  over- 
hanging portion  carries  a  vertical  load  Q.  The  length  BC 
equals  OB,  =2a.  Required  the  vertical  deflection  of  the 
point  D,  neglecting  the  work  of  shear  (that  due  to  thrust  being 

zero,  as  before).    By  eq.  (25)  this  deflection  y\  -jp,  where  U 

*See  Mr.  C.  W.  Hudson's  paper  on  "Deflection  of  Beams  of  Variable 
Moment  of  Inertia,"  in  Vol.  LI  (Dec.,  1903)  of  the  Transac-  Am.  Soc.  Civ. 
Engineers,  p.  1. 


54 


MECHANICS    OF    INTERNAL   WORK 


§41 


is  the  total  internal  work 


.    This  summation  will  be 


fc  M*dx 
'Jo    2EI 

separated  into  the  three  integrals  (reckoning  x  for  each  seg- 
ment in  the  way  shown  in  Fig.  18)  for  OD,  DB,  and  BCy 
respectively, 


-I— 


FIG.  18. 


In  this  case  we  note  that  from  ordinary  statics  the  reaction 

p 
at  0  is  V^-jr—  Q>  and  obtain  the  following  expressions  for  M 

and  its  various  derivatives  in  the  three  segments,  viz., 


=-e     x;  and  = 


For    OD,    M 
for  points  on  DB, 

M=V(a+x)-Px,=(^~Q)(a+x)-Px',     and 
while  on  CB,  M  =  Qx    and    ~J'=^ 


Before  substitution,  however,  we  may  write  (differentiating 
under  the  integral  sign), 


§  41  DEFLECTION    OF    BEAMS  55 

in  which,  wnen  values  are  substituted  from  above  relations  for 

M,  etc.,  there  results 


(84) 


Performing  the  integrations  and  reducing,  we  obtain 

[P       ~ 
j- 


In  this  equation,  if  we  now  make  Q  equal  to  zero,  we  have 
the  same  result  as  in  eq.  (78)  ;  while  if  P  be  made  equal  to  zero 
we  have  the  deflection  (upward;  note  the  negative  sign)  of  the 
point  D  as  due  to  the  single  load  Q  at  C  (in  which  case,  of 
course,  the  extremity  0  must  be  "  latched  down,"  and  the 
reaction  V  is  downward). 

It  should  be  carefully  noted  that  in  the  above  solution  P 
and  Q  are  independent  loads;  that  is,  when  P  is  conceived  to 
vary  Q  remains  constant;  in  other  words,  Q  is  not  a  function 
of  P  and  hence  (as  above)  d(Qx)/dP  =  Q.  But  the  reactions, 
V  and  V,  depend  on  both  P  and  Q;  and  neither  V  nor  V  can 
be  considered  constant  if  either  P  or  Q  is  to  vary,  but  must  be 
expressed  in  terms  of  the  variable  (P;  above)  before  dM/dP 
can  be  obtained  (on  OD  or  DB).  We  note  also  that  the  external 
work  done  by  these  two  reactions,  when  the  loads  P  and  Q  are 
gradually  applied,  is  zero;  since  the  supports  have  smooth 
horizontal  surfaces  and  the  points  of  the  beam  in  contact  with 
them  are  thus  free  to  slide  (one  or  both)  during  the  loading; 
thus  justifying  us  in  the  application  of  eq.  (25),  or  (19).  [See 
note  following  eq.  (19)]. 

In  case  the  vertical  deflection  (y2),  of  extremity  C  were 
desired,  there  being  a  vertical  load  Q  at  that  point,  we 
should  have 

»-;     ....    -    -  '•'.'  (86) 


the  expansion  of  which  would  give  us  eq.  (83)  with  P  replaced 
by  Q.  In  such  case,  of  course,  d(Qx)/dQ  is  not  zero,  but  =x. 
(Let  the  student  work,  out  the  detail  of  finding  y2.) 


56 


MECHANICS    OF    INTERNAL   WORK 


§42 


42.  Example  IV.  Equation  to  Curve  of  Straight  Beam. 
Two  End  Supports.  Single  Eccentric  Load.  It  is  proposed 
to  use  our  present  methods  for  determining  the  equation  of 
the  elastic  curve  OD  of  the  beam  OC  in  Fig.  19  (homogeneous, 


FIG.  19. 

prismatic,  and  originally  straight)  resting  on  two  end  supports 
(smooth,  horizontal  surfaces)  and  loaded  with  a  concentrated 
load  Q  not  in  the  middle  but  at  a  distance  a  from  the  left-hand 
support  greater  than  that,  6,  from  the  other  support.  (Denote 
a  +  b  by  I.)  We  shall  first  find  the  vertical  deflection  of  any 
point  B  (between  0  and  D)  whose  distance  from  0  is  x\,  by 
the  process  just  illustrated,  involving  the  use  of  a  "  dummy  " 
vertical  force  P  applied  at  B.  With  such  a  force  conceived  to 

act,  the  reactions  of  the  supports  are  evidently  Vo  =  jQ  + 


I 

and  Vc=jQ+-j-P,  respectively.     The  distance  of  any  point  n 
from  0  will  be  called  x,  and  the  moment  of  stress-couple  at  any 

while 


Again,  for  any  section  between  B  and  D,  we  have 


-P(x-xl);    and         - 


For  cross-sections  on  portion  DC  it  will  be  more  convenient 
to  reckon  x  (calling  it  now  x')  from  the  support  C;  whence, 
for  any.  section  between  D  and  (7, 


A 
and     —  =y 


§  42        EQUATION  TO  CURVE  OF  BENT  BEAM          57 


=-^T  I  M  \Jpj 


Now  the  deflection  of  the  point  B  is  y  I=-^T  I  M  \Jpjdx, 

the  summation  extending  over  all  the  dx's  (or  dxf  's)  of  the  axis 
of  the  beam  and  can  be  written  equal  to  the  sum  of  three 
integrals  [applying  to  the  segments  OB,  BD,  and  DC,  respect- 
ively], viz.  : 


•or  for  brevity, 

.....     (87) 


Before  substituting  details  in  eq.  (87)  we  may  put  P  =  0  in 
each  expression  for  M,  since  in  this  way  the  final  result  is  the 
same  as  if  the  P  had  been  retained  and  had  not  been  placed 
«qual  to  zero  until  the  final  expression  had  been  reached;  but 

-each  -jp-,  of  course,  is  not  equal  to  zero.     In  detail,  therefore, 
bearing  in  mind  these  substitutions,  we  obtain,  successively, 

bQ  Xl* 


that  is, 


while 

i-x'dx'=~r)0  *'w=  P  '•f- 

Since  B  is  arbitrary  in  position  we  may  regard  it  as  any 
point  whatever  on  the  elastic  curve  OD;  and  hence  x\  and  yi 
become  abscissa  and  ordinate  of  any  such  point.  Therefore, 
substituting  the  values  of  Z\,  Z2,  and  Z3  just  found,  in  eq.  (87), 


58  MECHANICS   OF   INTERNAL   WORK  §  43 

and  with  y  for  y\  and  x  for  x\  (for  simplicity),  we  have,  as  the 
equation  to  the  elastic  curve  OD,  0  being  the  origin, 


Q  y  r    "'  3 

Or,  after  reduction, 

IEI       [abl 


**»     '   •'"'   '   (   } 

as  the  equation  to  elastic  curve  OD. 

The  abscissa  of  the  lowest  point  of  this  curve  (note  that 
OD  is  >DC',  i.e.,  a>b)  which  will  thus  locate  the  point  of 
maximum  deflection,  may  be  obtained  by  writing 

dy  Q  \abl    a?b     bx^ 

I-0'  i.e,byputtingM[T-^-2-J=0; 

which  being  solved,  gives  for  the  required  abscissa  x,  —xm, 


which  can  also  be  written 


This  value  for   xm  being   substituted  in  eq.  (88)  gives  for 
maximum  deflection, 


-«(a-a)-  •  •  •  •  (90> 

or,  in  another  form, 


(These  results  for  xm  and  ym  are  seen  to  check  with  those  on 
pp.  258,  494,  and  506  of  M.  of  E.) 

43.  Examples  for  Practice.     Straight  Prismatic  Beams. 

The  following  involve  horizontal  straight  beams  of  constant 
moment  of  inertia: 

Example  V.  For  the  cantilever  beam  shown  at  (A)  in 
Fig.  20,  find  the  vertical  displacement  (i.e.,  deflection)  of  the 
extremity  N,  the  load  W  being  uniformly  distributed  over  the 


§43 


DEFLECTION    OF    BEAMS 


59 


whole  length;   w  is  the  load  per  running  foot  and  I  the  length; 
/  constant. 


Ans. 


'  El' 


Example  VI.  The  cantilever  beam  at  (B)  in  Fig.  20  is 
built  in,  horizontally,  at  the  left-hand  extremity  and  carries 
a  variably  distributed  load,  W]  the  rate  of  distribution,  w, 
is  proportional  to  the  distance  x,  from  the  free  end,  N.  Find 
the  vertical  deflection  of  point  N 


WP 
™  '  El' 


Ans. 


FIG.  20. 

Example  VII.  Find  the  vertical  .displacement  of  any  point 
on  the  elastic  curve  of  the  beam  at  (A)}  Fig.  20,  in  terms  of 
the  horizontal  distance  x\,  from  extremity  N. 


w 


Ans. 


Example  VIII.  Find  the  vertical  displacement  of  any 
point  on  the  elastic  curve  of  the  beam  at  (B),  Fig.  20,  in  terms 
of  the  distance  x\  from  the  extremity  N. 

w  r«  -~       .1 

Ans. 


44.  Horizontal  Straight  Beam  of  Variable  Moment  of 
Inertia  (i.e.,  a  Non-Prismatic  Beam).  Resting  on  Two 
Terminal  Supports.  As  a  case  presenting  this  feature  let  us 
take  the  beam  shown  in  Fig.  21.  All  its  sections  are  rectan- 
gular and  the  height  is  constant  throughout,  =h.  The  two 
extreme  quarters,  however,  are  wedge-shaped,  the  width  u 


60 


MECHANICS   OF   INTERNAL   WORK 


§44 


being  proportional  to  the  distance  x  from  the  end,  while  the 
middle  half  DA  is  prismatic,  of  constant  width  6.  A  single 
concentrated  load,  PI,  being  placed  at  the  middle  point  B,  it 
is  required  to  find  the  vertical  displacement  of  that  point;  or  y\. 
From  eq.  (25), 

dU         c  M    dM 


that  is,  E  being  constant, 
2  CBM    dM         2 


dM 


>~8~t  -. 


I 
FIG.  21. 

With  x  measured  from  support  0,  we  have,  for  the  dx's 
along  AB,  I  being  constant,  =/o, 

_.    PI        dM    x          ,     ,    bh?       . 
M=-^-x;    -7p-  =  «;     and     /=TO,  =  A>; 

Zi  dL    \         —  1- — 

while  along  OA  (/  variable), 


^i        am     x  un*    x    on*    x 

~5~x'>    ^P"=O»    an(i    **     =T9~==7T  •  T9"=^  •  ^°' 
air  i      J  LZ      Q,      LZ      & 


Hence  eq.  (92)  becomes 
2[  Ta  a     Fix    x 


L     Piz    x 

"  •  ~17~  •  77  •  < 

0         ^         •" 

STo  *     P7      ~     2^kiT3    *        WW 


45.  Horizontal  Non-Prismatic  Beam.  I  Constant  through- 
out Each  of  Several  Portions  of  the  Length.  (Fig.  22.) 
In  the  case  here  shown  the  cross-section  has  different  moments 


46 


STATICALLY    INDETERMINATE    BEAMS 


61 


of  inertia  (7)  in  the  different  portions  OA,  AB,  etc.,  of  the 
length  (an  I-beam  with  strengthened  flanges,  for  instance),  but 
the  value  of  /  is  constant  through  any  one  such  portion.  With 
a  symmetrical  arrangement  as  indicated  in  Fig.  22,  and  with  I\ 
denoting  the  moment  of  inertia  along  AO  and  CE,  72  that  along 
AB  and  DE,  and  73  along  BH  and  HD,  we  obtain  yi  of  the 
middle  point  H  by  using  a  relation  like  that  in  eq.  (92)  but 
with  three  terms  in  the  bracket  instead  of  two;  and  in  the 
three  integrations  concerned  a  different,  but  constant,  7  would 
occur  in  each.  (See  foot-note  in  §  41.) 


'• 

1 

1 

°l 

/ 

t 

B                 H                | 

D                    1 

1°    , 

1 

1 

Ed 

| 

—,-4 

1 

1 

i  | 

FIG.  22. 

46.  Straight  Horizontal  Beams  on  More  than  Two  Sup- 
ports (Statically  Indeterminate).  If  the  supported  points 
of  the  beam  are  all  free  to  slip  along  smooth  fixed  surfaces 
(except  that  one  point  must  be  fixed)  and  there  are  more  than 
two  points  supported,  the  beani  is  statically  indeterminate;  i.e. 
the  reactions  (and  consequently  the  internal  stresses)  cannot 
be  determined  by  simple  statics.  Our  present  methods,  how- 
ever (Castigliano's  Theorem  and  the  Theorem  of  Least  Work), 
enable  solutions  to  be  obtained  by  very  simple  and  straight- 
forward steps.  A  beam  on  more  than  two  supports  is  called 
a  "  continuous  girder." 

It  must  be  recalled  that,  in  accordance  with  Castigliano's 
Theorem  (§  21),  if  a  beam  is  supported  at  two  points  (one  fixed; 
the  other  free  to  slide  on  a  smooth  fixed  surface)  the  displace- 
ment y\  of  the  point  of  application  of  some  one  load  PI,  con- 
sidered as  a  variable,  in  the  direction  of  pointing  of  this  load, 

will  be  given  by  2/i=-Tp-,  provided  each  of  the  other  external 
forces  either  is  independent  of  P\,  or,  on  account  of  being  the 


62  MECHANICS    OF    INTERNAL    WORK  §  47 

reaction  of  a  smooth  fixed  support,  does  no  external  work  when 
the  beam  is  loaded. 

If,  therefore,  a  beam  has  two  points  supported  and  is  to 
carry  several  loads,  P,  Q,  etc.,  and  we  desire  that  the  point 
of  application  of  some  one  of  them,  say,  Q,  shall,  during  the 
gradual  loading  of  the  beam,  execute  no  movement  whatever 
in  the  direction  of  Q;  in  other  words,  that  its  displacement 

shall  be  zero;  we  have  simply  to  write  — rr  =  zero  and  solve  the 

equation  for  Q.  That  is,  if  the  point  of  application  of  Q  is 
prevented  from  moving  in  the  direction  of  Q,  during  the  loading 
of  the  beam,  by  bearing  against  a  fixed  smooth  surface  at 
right  angles  to  the  direction  of  Q,  the  reaction  of  that  surface 

7TT 

will  be  Q,  and  must  have  the  value  computed  from  -^  =  0. 

d*t 
Hence  if,  before  loading,  the  point  of  application  of  Q  be 

chosen  as  a  third  point  of  support  for  the  beam  (with  a  fixed 
bearing  surface  provided  at  right  angles  to  Q)  the  value  of  the 
reaction  after  loading  may  be  found  as  indicated. 

It  is  evident  that  the  same  reasoning  applies  to  any  elastic 
structure  (without  initial  stresses)  as  well  as  to  a  beam,  and 
also  applies  when  there  are  more  than  two  points  of  the  beam 
or  structure  already  provided  with  supports  (of  the  nature 
indicated  above) ;  but,  in  any  case,  all  the  external  forces  which 
are  dependent  on  the  unknown  Q  must  be  expressed  as  functions 

of  Q  before  the  derivative  in  question  is  taken;    then  -377  =  0 

will  give  Q  in  terms  of  other  external  forces. 

47.  Continuous  Girder.  Case  I.  (Fig.  23.)  A  straight, 
prismatic,  homogeneous  beam  C'OC  is  placed  on  three  unyield- 
ing equidistant  supports,  at  the  same  level.  The  two  spans 
C'O  and  OC,  equal  in  length,  each  length  being  Z  =  a  +  6,  are 
loaded  symmetrically  with  two  loads,  each  =  P;  one  in  each 
span,  at  distance  a  from  the  middle  support. 

This  beam  is  a  "  continuous  girder  "  for  which  at  the  outset 
the  reactions  of  the  three  supports  are  unknown.  The  reactions 
at  C  and  C"  are  equal,  from  symmetry,  each  =  F;  that  at 
support  0  is  V.  They  cannot  be  found  by  ordinary  statics 
and  it  is  now  proposed  to  determine  them  by  the  principle 
announced  in  the  last  paragraph. 


:§47 


CONTINUOUS    GIRDERS 


63 


Selecting  V  as  the  reaction  to  be  found  by  the  method  of 
§  46  we  are  to  put  377  =  0-     For  the  present  beam  there  is  no 

thrust,  and  the  work  of  shear  will  be  neglected;   so  that,  by 

.  (74)  and  (75),  §  38,  putting  dx  for  ds, 


dU      d 


1 


dM 


extended  over  the  whole  beam.  The  moments  (M)  of  the 
stress-couples  along  the  beam  will  aU  be  obtained  in  terms  of 
the  reaction  V,  at  C  (regarded  as  a  variable),  and  of  the  known 
(constant)  load  P.  On  account  of  symmetry  it  will  only  be 
necessary  to  deal  with  the  internal  work,  C7,  of  the  right-hand 


half,  OC',  of  the  beam;  the  expression  involving  which  can  then 
be  doubled,  for  the  total  internal  work. 

For  any  dx  on  the  portion  DC  (the  abscissa  x  being  meas- 
ured as  in  the  figure)  we  have 

M  =  Vx-,    and    'jy'==x> 

while  for  any  dx  on  OD  (and  for  these  dx's  we  reckon  x  from 
point  D), 


M  =  V(b+x)-Px;    and    jr- 
Hence,  substituting  in  eq.  (94),  we  obtain 


'     (95) 


64 


MECHANICS    OF    INTERNAL    WORK 


§48 


The  integration  being  performed  and  the  limits  inserted, 
there  results 


2' 


(When  once  V  is  found    the  other  reactions  easily  follow  from 
ordinary  statics)  (viz.,  at  C',  V;  and  at  0,  2P-2V). 
In  the  particular  case  where  b  =  a  this  gives 


(97) 


and  hence 


F'--  tf-P. 


48.  Continuous  Girder.  Case  II.  (Fig.  24.)  Straight, 
prismatic,  homogeneous  beam  on  three  equidistant  and  unyield- 
ing supports,  0,  B,  and  C,  at  same  level.  There  is  but  one 


FIG.  24. 

load,  viz.,  P,  in  middle  of  left-hand  span.  Evidently  the 
support  at  0  must  be  placed  above  the  beam  and  the  reaction, 
Vj  at  that  extremity  must  point  downward,  since  there  is  no 
load  between  0  and  B.  V  will  be  selected  as  the  variable. 
The  beam  must  be  divided  into  three  portions  OB,  BD,  and 
DC  for  summation  purposes,  but  first  the  other  two  reactions, 
V  and  V",  since  they  depend  on  V,  must  be  expressed  as  func- 
tions of  V.  This  is  done  by  ordinary  statics,  the  whole  beam 
being  the  "  free  body";  first  taking  moments  about  C,  and 
then  about  B',  whence 

F'  =  27  +  iP    and     V"  =  \P  -V  .    .     .     .     (98) 

Reckoning  x  as  shown  in  figure,  for  the  three  respective 
portions  of  beam,  we  have 


onOJ5: 


and 


§49 


i.e., 


CONTINUOUS   GIRDERS 


,65 


and 


dM 
dV 

dM 


on  DC:    M=F"£=(iP-F)z;    and 
Hence,  substituting  in  eq.  (94), 

f\V(2a-x)-%Px](2a-x)dx 

JQ 


whence,  solving, 


'  =        and 


49.  Continuous  Girder.  Case  III.  Non-Prismatic 
Beam.  The  straight  homogeneous  beam  of  Fig.  25  is  sup- 
ported at  three  equidistant  points  at  same  level.  Supports 
firm.  Two  equal  loads,  each  =  P,  one  in  middle  of  each  span. 


FIG.  25. 

The  middle  half  DA  is  prismatic,  having  a  rectangular  section 
with  constant  moment  of  inertia,  /o,  =bh3-t-12.  The  extreme 
quarters,  however,  CD  and  AO,  have  rectangular  sections  of 
constant  height  h  and  variable  width  u,  u  being  proportional 
to  distance  x  from  0,  or  (7;  i.e.,  u=  (x+a)b,  so  that  the  moment 
of  inertia  along  these  portions  of  the  beam  is  variable  and  is 
I  =  uh*  +  l2  =  (x  +  a)I0. 

The  unknown  reactions  being  V,  V,  and  V,  if  V  be  chosen 

as  the  variable  its  value  is  found  by  placing  -^  I     -j-. 


66 


MECHANICS   OF   INTERNAL   WORK 


§50 


the  work  of  shear  being  neglected  and  advantage  being  taken  of 
symmetry. 


OnOA,     M=Vx-        r  =  x;    and    7  =   /0. 


On  AB,     M=Vx-P(x-a};    j^  =  x;    and     I  =  I0. 
Hence 


2o 


Va* 


7Pa*    3Pa3 


.    ,  (98a) 


=  AP;    and     7'  =  ffP (986) 


and 


50.  Continuous  Girder.  Case  IV.  Symmetrically 
Placed,  with  Uniform  Loading.  (Fig.  26.)  As  in  the  preced- 
ing cases  (except  Case  III),  the  beam  is  prismatic  (I  constant 
along  whole  length),  with  two  equal  spans;  supports  on  a 


Iv  |V  v± 

lUiuiiuliiiimiul 


level;  but  the  load  is  uniformly  distributed  over  the  whole 
length,  at  rate  of  w  Ibs.  per  foot.  (Fig.  26.)  The  load  on 
one  span  is  W  =  wl,  and  the  reactions  are  V,  V  and  V.  As 
before,  taking  F  as  a  variable,  and  doubling  the  result  obtained 
from  a  consideration  of  the  dx's  along  one  span  0(7,  or  half 
the  length  of  beam,  we  are  to  put 


2  C 
Eljc 


§  51  CONTINUOUS    GIRDERS  67 

The  M  for  any  dx  is  the  same  function  of  x  for  all  points  of 
CO;  i.e.,  from  the  free  body  of  length  x,  we  have  along  CO, 

x     T7      wx2  dM 

M=Vx—wx.-==Vx  —  ~;    and    ~Jy'==x 

w 


VI3     wl* 

--  --       =  0;     or     V  =  lwl  =  lW,    and     V'  =  *fW.  .    (98c) 


51.  Continuous  Girder.  Case  V.  (Fig.  27.)  Beam  pris- 
matic, and  loaded  uniformly  over  whole  length.  Three  points 
of  support  at  same  level.  Length  of  right  span  double  that  of 

1v  ±v'  V"A 

wa  2wa 

..  miimulmuumumiiuuiL 


FIG.  27. 


the  left.     Rate  of  loading,  w,  the  same  throughout  the  whole 
length,  which  is  3a.     E  and  7  being  constant,  we  may  write, 


(omitting  the  outside  factor  I/  El),  if  V  be  taken  as  a  variable; 
the  other  reactions  being  V  and  V"  .  Measuring  x  for  the 
dx's  on  OB,  and  BC,  as  shown  in  the  figure,  respectively,  we 
have  for  points  on  OB: 


.,     T7 

M=Vx  —  ^-     and     -Tf7  =  £- 
2  aK 

Taking  a  free  body  of  length  x  measured  from  (7,  we  obtain, 
for  points  on  span  BC,  M=V"x—wx2;  but  since  a  variation 
in  V  would  involve  a  variation  of  V"  also,  we  must  express  V" 
as  a  function  of  V  before  proceeding  further.  This  is  done, 
with  the  whole  beam  as  a  free  body,  putting  by  £(moms.)==0, 
about  B:  whence 

V    3wa 
~~      ~~ 


68  MECHANICS    OF    INTERNAL   WORK  §  52 

and 

TUT       T*r     (V3wa\       wx*  dM    x 

.-.     M  on  £C=^_+_Jz  -_;    and    w=^. 


Hence,  substituting  above, 
f«/T,      wx2\ 

Jo  (F*-—  )xdx+ 

i.e., 


Q 

Fa3    tm4     27a3 


383 
A     F  =  ^;     V'  =  %$wa;    and     V"  =  }%wa.   .    .     (99) 

The  shear  and  moment  can  now  be  computed  for  any  cross- 
section  of  the  beam. 

52.  Derivative  of  Internal  Work,  U,  with  Respect  to  MI, 
the  Moment  of  the  Stress-Couple,  at  Any  Section  of  a  Beam 
in  Flexure,  when  Statically  Indeterminate.  At  any  cross- 
section  (1)  of  a  beam,  straight  or  curved,  if  we  imagine  the 
connection  of  the  portions  of  the  beam  meeting  at  this  section 
to  consist  of  a  hinge  joint  at  the  center  of  gravity  of  the  cross- 
section,  and  two  short  bars  (one  on  each  side  of  the  hinge, 
and  at  a  proper  distance  from  it)  parallel  to  the  axis  of  the 
beam  at  that  point,  and  a  distance  a  apart;  then  the  pressure 
on  the  hinge  joint  will  be  the  resultant  of  the  thrust,  T\,  and 
shear,  Ji,  at  that  section  of  the  beam,  while  the  stresses  PI 
and  PI'  in  the  two  bars  will  be  equal  to  each  other,  and  the 
product  P\a  is  the  moment,  MI,  of  the  stress-couple. 

Now  if  the  beam  is  so  supported  as  to  be  statically  indeter- 
minate and  section  1  is  situated  between  two  supports  the  two 
short  bars  above  mentioned  might  be  omitted  and  the  beam 
would  still  remain  in  place.  In  other  words,  each  of  the  two 
bars  is  redundant;  and  we  therefore  write 

dU  dU 

_  =  0,     and    ^-0; 

that  is,  dividing  by  a,  and  adding, 

du       "Vo doo) 


§  53  THEOREM    OF   THREE    MOMENTS  69 

But  this  is  simply  the  complete  derivative  of  U,  the  internal  work 
of  the  whole  beam  with  respect  to  the  moment,  MI,  at  cross-section 
1.  (The  internal  work  of  the  two  short  bars  themselves  is 
considered  zero;  in  other  words,  they  are  considered  to  be 
infinitesimal  in  length.)  We  may  therefore  write 

' 


This  relation  is  available  for  determining  the  moment  of 
couple  MI,  at  any  given  section  (1)  of  a  statically  indeterminate 
beam,  straight  or  curved  (in  the  latter  case  the  radius  of 
curvature  must  be  relatively  large.) 

53.  Continuous  Girders.  Theorem  of  Three  Moments. 
Although  Castigliano's  Theorem  may  still  be  used  for  deter- 
mining the  support  reactions  in  case  a  continuous  girder  rests 
on  more  than  three  supports,  the  detail  is  very  burdensome  and 
tedious.  The  relation  just  proved  in  §  52,  however,  may  be 
brought  into  play  with  great  advantage  in  such  cases,  the 
treatment  being  very  much  simplified  thereby.  The  beam 
being  assumed  to  be  homogeneous,  straight,  and  prismatic 
(constant  /),  and  all  the  supports  to  be  at  the  same  level  and 
unyielding,  a  general  relation  based  on  eq.  (101)  will  first  be 
proved,  known  as  Clapeyron's  Theorem  of  Three  Moments, 
established  as  follows  : 

In  Fig.  28  we  have  a  straight  continuous  beam  with  three 
points  of  its  axis,  viz.,  0,  1,  and  2  (0  and  2  being  at  the  ends) 
supported  on  fixed  supports  at  same  level  (two  of  these  points 
are  free  to  slide  on  smooth  horizontal  surfaces).  Various 
vertical  loads,  P,  P,  etc.,  act  on  the  beam;  some  in  each  span; 
of  arbitrary  values.  Two  couples  are  also  applied  to  the  beam, 
one  at  each  end,  of  moments  MQ  and  M2,  respectively;  that  at 
the  right  being  clockwise;  that  at  the  left,  counterclockwise. 
These  couples  are  virtually  arbitrary  "  loads,"  like  the  P's, 
being  independent  of  them  and  of  each  other.  It  is  plain  that 
whatever  value  might  be  assigned  to  any  one  of  these  "  loads," 
the  beam  would  still  be  completely  supported.  Evidently  the 
reactions  V0,  Vi,  and  V2,  at  the  points  of  support,  are  each 
vertical. 


70 


MECHANICS    OF    INTERNAL   WORK 


§53 


For  any  set  of  given  values  for  the  "  loads  "  there  will  be  a 
corresponding  value,  MI,  of  moment  of  (internal)  stress-couple 
at  cross-section  1,  over  the  intermediate  support  at  1.  The 
beam  being  statically  indeterminate)  MI  may  be  found  by  the  use 
of  eq.  (101);  after  rewriting  it  (neglecting  work  of  shear)  thus: 


du     d 


2EI 


•tfMfSfy*  •  aw 


For  sections  on  span  01  x  is  reckoned  from  0;  for  those  oa 
span  02,  from  2.  For  section  n  on  01,  taking  On  as  "  free  body  " 
and  n  as  center  of  moments,  we  have,  for  the  moment  M, 


(103) 


where         (Pa/)  denotes  the  sum  of  the  moments,  about  n,  of  all 

^4  n 

the  loads  P  situated  between  n  and  0.  But  VQ  is  a  function  of 
MI,  obtained  by  considering  01  as  free  body  and  taking  moments 
about  point  1,  viz.: 


(104) 


Here^jP  (Pa;")  denotes  the  sum  of  the  moments,  about  point 

1,  of  all  the  loads  P  between  1  and  0. 

Now  note  that  if  the  portion  01  of  the  beam  were  cut  apart 
from  the  remainder  and  set  up  horizontally  on  two  supports  at 
its  extremities  0  and  1,  and  loaded  with  the  same  loads  P  as 
before  in  same  positions  (but  vnthout  the  couples  acting  at  0 


§  53  THEOREM    OF    THREE    MOMENTS  71 


and  1)  the  reaction  at  the  support  0  would  be  R0=  — 

a\ 

and  the  moment  of  couple  at  any  section  n  would  be 

.....    (105) 


This  ideal  moment,  Mn,  which  would  be  occasioned  at  any  sec- 
tion n  if  the  portion  01  were  supported  at  its  extremities  as  a 
"  simple  beam  "  [bearing  the  same  loads  P,  but  without  the 
end  couples]  may  be  called  the  "  normal  moment  "  (as  on  p.  494, 
M.  ofK). 

As  a  result  of  substituting  the  expression  for  R0  in  eq.  (105) 
and  combination  with  eqs.  (103)  and  (104)  we  readily  derive 
for  a  point  n  in  the  left-hand  span  0  ...  1, 

M  =  M0-M»-(M0-Afi)^-,      .     .     .     (106) 
giving  M  as  a  function  of  the  one  variable  MI  ;  whence 

3IT--.    il    •    -.    •    •    d07) 
dMi     a\ 

Similarly,  there  are  obtained,  for  any  point  n  in  the  right  span 
1...2, 

M  =  M2  -Mn  -  (M2  -Mi)—,  (108) 


and  —-.    •     -    •    •     >    »     •     (109) 

a2 


Substituting  in  eq.  (102),  which  may  be  written 

dx  =  0,.     (110) 


we  obtain 

^-j     \M0x-Mnx-(M0-Ml)^-\dx 

\  =0.        (Ill) 

i  Ca>i\  x2^ 

+— I     \M2x-Mnx-(M2-Ml)- 

6^2     0      I  ^2 

Of  the  quantities  «i,  «2,  -M0,  MI,  M2,  and  Mn,  only  Mn 
is  a  function  of  x  in  each  span;  this  ideal  moment,  or  "  normal 
moment,"  having  the  meaning  just  given.  Performing  the 


72  MECHANICS    OF    INTERNAL   WORK  §  54 

integrations,  as  far  as  it  can  be  done  in  the  general  case,  we 
have,  after  reduction, 

6  Cai  6  fa* 

=—  )    Mnxdx  +  —  I    Mnxdx,   (112) 


which  is  Clapeyron's  Theorem  of  Three  Moments,  giving  a 
relation  between  the  three  moments  MO,  MI,  M2,  of  the  couples 
at  the  three  points  0,  1,  and  2;  of  a  homogeneous,  prismatic 
beam  supporting  a  set  of  vertical  loads  P,  P,  etc.;  the  three 
points  in  question  resting  on  three  fixed  supports  at  same 
level. 

Note  A.  It  must  be  carefully  noted  that,  of  the  integrals 
in  the  right-hand  member  of  (112),  the  first  one  deals  with  the 
left-hand  span  and  its  loads,  x  being  measured  from  its  left  end, 
or  point  0;  while  the  second  relates  to  the  right-hand  span, 
x  being  measured  from  its  right  end  or  point  2;  and  the  ideal 
Mn  must  be  a  positive  number. 

Note  B.  The  three  moments,  M0,  M\,  and  M2,  have  been 
used  in  such  a  way  as  to  imply  tension  in  the  upper  fibers  of 
the  beam,  and  results  must  be  interpreted  on  this  basis;  that 
is,  if  a  negative  number  is  obtained  for  one  of  them  it  implies 
that  at  that  section  of  the  beam  the  upper  fibers  are  in  com- 
pression. 

54.  Values  of  the  Two  Integrals  in  the  Three-Moment 
Theorem  in  Particular  Cases.  For  brevity,  the  first  integral 
(referring  to  loads  "in  the  left-hand  span)  in  eq.  (112)  may  be 
denoted  by  Zi,  and  the  second  (right-hand  span)  by  Z2;  and 
the  theorem  will  then  read 

A  f\ 

M0ai  +2M  1(01  +  a2)  +  M2a2=-  .  Zl  +-  .  Z2.    .    (113) 

a  0,2 

If  the  loading  on  the  left-hand  span  is  a  uniformly  distributed 
load,  Wi  covering  the  whole  span  (see  Fig.  28)  a\  at  rate  w\ 
Ibs./lin.ft.,  the  value  of  Mn  at  any  point  n  in  portion  01  [con- 
ceived to  be  a  separate  beam  resting  on  two  supports  at  its 


W\         i 
extremities]  would  bo  —x  --  —  ;  hence  (with  W  = 


§  54  THEOREM    OF    THREE    MOMENTS  73 

Similarly,  if  the  right-hand  span  carries  a  load  W2,  =  w2a2, 
•covering  the  whole  span  a2)  we  have 


'2=24 


TF2a23. 


With  a  single  concentrated  load  P  in  the  left-hand  span 

Px" 

(Fig.  28)  at  a  distance  x"  from  support  1,  we  have  Mn=  --  x 

a\ 

for  any  point  n  on  the  left  of  P,  but  for  any  point  n  between 

Px" 

P  and  support  1,  Mn  =  -  x  —P[x  —  (a\  —  x"}}  ;  hence 

tti 

,-ai-z"   /P:r"r\  rai          /  P'Vr 

Zl=  ^LfJ-Hfc+r  (£       ^-P[;C-(0l-x 

~/0  \     al     /  */ai-x"     \       al 

i.e., 

s"X';       -   •'..     -     (H5) 


and  similarly,  for  a  single  concentrated  load  P  in  the  right- 
hand  span,  at  distance  x"  from  support  1, 


(If  for  ai  —  x"  we  write  z"  (distance  of  P  from  support  0) 
Zi  in  (115)  becomes 

JP(ai2-z"2X';    and    Z2  =  JP(o22-2//2X/, 

for  a  load  P  in  right-hand  span,  with  z"  =  a2—x".) 

For  any  number  of  concentrated  loads  in  the  left-hand 
span,  Z\  is  the  sum  of  a  number  of  terms  of  the  form  given  in 
the  right-hand  member  of  eq.  (115),  one  term  for  each  load, 
and  containing  the  amount,  P,  of  the  load  and  the  correspond- 
ing x"  .  A  similar  statement  may  be  made  for  Z2,  as  regards 
the  right-hand  span,  if  that  span  carries  a  number  of  concen- 
trated loads;  remembering  that  the  x"  is  measured  from  support 
1  (see  Fig.  28). 

Of  course,  if  there  is  no  load  on  the  left-hand  span,  Z\  is 
zero;  and  similarly  for  the  right-hand  span,  with  no  load  on 
that  span,  Z2  =  0. 

For  a  distributed  load  over  any  portion  of  the  left-hand 
span,  the  rate  of  distribution  w  being  constant  or  variable,  we 
may  find  the  value  of  Zi  by  substituting  wdx"  for  P  in  eq.  (115) 


74  MECHANICS   OF   INTERNAL   WORK  §  54 

and  summing  (i.e.,  integrating)  all  such  terms  between  the 
extremities  of  the  distributed  load;  that  is,  by  making  x" 
vary  between  proper  limits.  If  to  is  variable  it  must  first  be 
expressed  in  terms  of  the  variable  x?'.  For  example,  if  a  load 
Wi  of  constant  rate  Wi  is  distributed  over  the  part  of  the  left- 
hand  span  which  extends  c\  ft.  toward  the  left  from  support  1 
(so  that  Wi=wid),  we  have 

«/x"=0 


and  similarly  if,  in  the  right-hand  span,  a  uniformly  distributed 
load,  PP2.  of  rate  u>2,  extends  to  the  right  over  c2  ft.  from  the 
support  1,  we  have 


Again,  if,  in  the  left-hand  span,  the  uniformly  distributed 
load  Wi,  =widi,  of  rate  w\,  extends  d\  ft.  toward  the  right 
from  the  support  0,  by  a  similar  process  we  obtain  (now  making 
z"  vary  between  the  limits  GI  —  di  and  ai), 


while  a  distributed  load  Wz,  =1020*2,  at  constant  rate  u>2,  cover- 
ing the  portion  of  right-hand  span  extending  toward  the  left 
d2  ft.  from  support  2,  calls  for  a  value  of  Z2=  ^WzdzQa**  —d#), 

If  neither  end  of  a  distributed  load  adjoins  a  support,  results 
for  Zi  and  Z2  are  obtained  by  the  same  methods,  by  simply 
using  proper  limits  for  the  variable  x";  (and  if  to  is  variable  it 
must  first  be  expressed  as  a  function  of  x"). 

In  conclusion  it  may  be  said  that,  for  any  land  of  (vertical) 
loading  whatever,  on  either  span,  the  value  of  Z\  (or  Z2),  is 
made  up  of  the  sum  of  the  separate  terms  that  would  apply 
to  each  separate  part  of  the  loading,  concentrated  or  distributed, 
carried  by  the  beam  in  the  span  hi  question. 

55.  Continuous  Girder:  (Prismatic)  Resting  on  an  Indefi- 
nite Number  of  Fixed  Supports  at  the  Same  Level.  In  such  a 
case  it  is  convenient  to  let  the  moments  hi  the  sections  just  over 


§  55  CONTINUOUS   GIRDERS    BY    "  THREE    MOMENTS  75 

the  various  supports  constitute  a  set  of  unknown  quantities 
to  be  deterimned  at  the  outset,  use  being  made  of  the  Theorem 
of  Three  Moments,  just  proved  in  §  53. 

To  apply  this  method  we  must  note  that  the  beam  in 
Fig.  28  may  be  considered  to  be  a  portion  (isolated  as  a  "  free 
body  "),  comprising  any  two  consecutive  spans,  of  a  long  con- 
tinuous girder  of  several  spans.  At  0,  a  section  has  been  made 
cutting  close  on  the  right  of  a  support,  MQ  being  the  moment 
of  the  couple  acting  in  the  section,  while  VQ  may  represent  the 
shear  in  the  section;  while  2  is  a  section  made  close  on  the 
left  of  a  support  (the  second  support  from  the  other),  M2  being 
the  moment  of  couple  and  Fo  representing  the  shear  in  the 
section.  We  have  therefore  only  to  apply  this  relation  to  as 
many  lengths,  of  two  consecutive  spans  each,  as  circumstances 
permit,  in  order  to  obtain  as  many  equations  as  there  are 
unknown  quantities. 

For  example,  if  there  are  five  spans,  i.e.,  six  points  of 
support  at  same  level  (the  beam  simply  resting  on  each  sup- 
port; not  clamped  or  "  built-in  ")  the  moments  over  the  six 
points  of  supports  are  the  quantities  MQ  (over  the  extreme 
left  support;  and  MQ  is  either  zero  if  the  beam  terminates  there 
or  there  is  an  unloaded  overhang;  or  is  already  known  if  there 
is  a  loaded  overhang),  MI,  M2,  M9,  M4,  and  finally  M5 
(which  is  either  zero  if  there  is  no  overhang,  or  an  unloaded 
overhang;  or  is  already  a  known  quantity  if  there  is  a  loaded 
overhang).  Here  the  theorem  may  be  applied  four  times: 
First  to  the  two  consecutive  spans  extending  from  support  0 
to  support  2 ;  then  from  1  to  3 ;  then  from  2  to  4 ;  and  finally 
from  3  to  5.  The  equations  thus  obtained  are  seen  to  be  four 
in  number,  containing  no  unknown  quantities  except  the  four 
moments,  MI  to  M4  inclusive,  which  are  then  determined  by 
ordinary  algebra.  With  these  moments  known,  all  reactions, 
and  the  moment  and  shear  at  any  section,  are  easily  found  by 
ordinary  statics. 

In  case  the  left-hand  end  of  the  beam  is  clamped,  or 
"  built-in/'  horizontally,  we  may  consider  that  the  first  actual 
span  on  the  left  is  the  right-hand  span  of  two  consecutive 
spans  of  which  the  left-hand  span  is  indefinitely  short  (ai=0) 
and  carries  no  load;  so  that  the  theorem,  being  applied  to 


76  MECHANICS    OF    INTERNAL    WORK  §  56 

these  two  spans  gives  a  result  obtained  by  putting  a\  and  Z\ 
each  equal  to  zero  in  eq.  (113), 

2Mia2  +  M2a2  =  (6-ba2)Z2.    .     /'--:,     (118) 

Similarly,  if  the  right-hand  extremity  of  the  girder  is  fixed, 
or  built-in,  horizontally,  we  consider  the  last  actual  span  on 
the  right  to  be  the  left-hand  span  of  two  consecutive  spans  of 
which  the  right-hand  one  is  indefinitely  short,  and  carrying  no 
load;  that  is,  we  put  a2  and  Z2,  each  equal  to  zero  in  eq.  (113), 
and  obtain 

M<>ai+2Afiai  =  (6^ai)Zi.    .-.,,.     .     (119) 

Of  course  in  both  of  these  special  cases  the  built-in  section 
of  the  beam  is  represented  by  the  section  at  point  1  in  Fig.  28; 
with  corresponding  meaning  of  the  symbols  in  the  above  two 
equations. 

56.  Continuous  Girders.  Case  VI.  (Fig.  29.)  This  figure 
shows  a  continuous  girder  extending  over  three  spans  and  with 
its  extremities  resting  (not  built-in)  on  the  extreme  supports, 


f° 

wa 

liiUUUU 

II 

wa 

IJIIUi 

ilijiiaiiiimiiiiiiiiii! 

oj 

i 

a  

f 

(\}2                                              "31 

j  I 

1  u 

ft            2a- 

i  ! 

0  and  3.  The  two  spans  on  the  left  are  each  of  length  a  and 
carry  a  uniformly  distributed  load  (of  the  same  rate,  w,  per 
lin.ft.  on  both  spans)  throughout  their  entire  lengths.  The 
other  span  has  double  the  length  of  either  of  the  first  and 
carries  a  uniformly  distributed  load  over  its  whole  length,  of 
rate  w'. 

The  girder  is  homogeneous  and  prismatic  and  the  four 
supports  are  at  same  level,  hence  the  Theorem  of  Three  Moments 
is  available  [eqs.  (112)  and  (113)].  Applying  it  first  to  the 
two  consecutive  spans  on  the  left,  viz.,  0  ...  1  and  1 ...  2,  we 
find  [see  eq.  (114)],  with  M0  =  0, 


§  56  CONTINUOUS    GIRDERS    BY    "  THREE    MOMENTS  "  77 

Next  applying  it  to  the   (consecutive)   spans   1  ...  2  and 
2  ...  3,  we  have,  with  M3  =  Q, 

6    (wa)a*  .   6     (2ti/q)(2o)« 
-.-^-+^.-  -^-  - 

.     .  .\    .     (121) 


By  elimination  between  (120)  and  (121)  we  have 
Mi  =Hwa2  -^w/a2;     and     M2  =  A-i 


Assuming  that  w  is  greater  than  T8rW,  we  note  that  both  MI 
and  M2  are  positive;  that  is,  that  the  couples  acting  in  the 
sections  at  1  and  2  are  of  a  character  producing  tension  in  the 
upper  fibers  at  those  sections  (according  to  Note  B  of  §  53). 
In  other  words,  the  curvature  of  the  beam  is  convex  on  the 
upper  side;  in  the  way  indicated  at  the  points  0,  1,  and  2, 
in  Fig.  28. 

The  reactions  VQ,  V\,  etc.,  are  now  easily  found  by  statics. 
For  example,  from  the  free  body  0  ...  1,  reaching  from  sup- 
port 0  to  a  cutting  plane  just  short  of  support  1,  we  find  acting 
at  the  right-hand  end  of  the  body  a  shear,  J'  ',  and  a  couplo 
(clockwise)  whose  moment  is  M\,  just  obtained.  Putting 
£(moms.)=0  about  the  neutral  axis  of  this  section  for  this 
body  we  find 

J'  X  0  -  V0a  +  wa  .  £  -  Ml  =  0, 
and  hence  V^=^wa+-^w'a,      .....     (122) 


and  similarly  we  may  find  the  other  three  reactions,  taking 
various  "free  bodies";  the  whole  beam  being  conveniently 
taken  as  such,  for  two  of  them. 


78  MECHANICS   OF   INTERNAL   WORK  §  57 

CHAPTER  V 
COMPOSITE  SYSTEMS;   AND  CURVED  BEAMS 

57.  Straight  Prismatic  Beam  Supported  by  Three  Others. 
Symmetrical  Design.  In  Fig.  30  we  have  a  system  of  four 
horizontal  straight  beams,  each  prismatic  and  homogeneous;  the 
upper  one,  A,  resting  on  the  other  three,  B,  B',  and  B,  under- 
neath; and  these  latter,  on  unyielding  supports,  C"  and  C". 
A  is  at  right  angles  to  the  others,  touching  the  middle  of  each. 
The  middle  of  A  touches  B'\  and  its  two  extremities  touch 


FIG.  30. 

B  and  B.  For  A  we  have  the  values  E  and  7;  for  each  of  the 
other  beams,  E'  and  /'.  There  is  but  one  load,  viz.,  P,  vertical 
and  applied  at  the  middle  of  A.  Weights  of  beams  neglected. 
Before  the  system  is  loaded  the  beams  are  in  dose  contact  with 
each  other  and  with  the  supports,  without  strain;  Castigliano's 
Theorem  is  therefore  applicable;  and  its  corollary  (Least 
Work). 

The  nine  pressures  between  the  beams  and  against  the 
supports  are  all  unknown  in  advance  and  cannot  be  deter- 
mined by  statics  alone,  but  they  can  all  be  expressed  in  terms 
of  one  of  them,  say,  7,  the  pressure  under  the  left-hand 
extremity  of  beam  A.  That  at  the  right  extremity  of  A  is 
also  V]  and  that,  V,  under  the  middle  of  A  is  readily  seen 


I  57  COMPOSITE    SYSTEMS  79 

to  be  equal  to  P—  2V,  for  the  equilibrium  of  /L.  Under  each 
•  end  of  beams  B,  B,  the  pressure  is  V/2.  Under  each  end  of 
B'  it  is  772;  that  is,  (P-27)/2.  The  various  half  lengths 
are  a  and  6;  while  the  x  of  any  point  is  measured  as  in  Fig.  30. 
If  we  consider  the  left  end  of  beam  A  to  bear  on  beam  B 
by  the  interposition  of  an  infinitely  short  vertical  bar  the  stress 
in  this  bar  will  be  the  unknown  7,  and  we  may  take  this  bar 
as  the  redundant  bar  of  the  system;  and  hence  (by  §  29)  write 
dU/dV  =  0  (the  internal  work  of  the  bar  itself  being,  of  course, 
zero).  Here  U  must  denote  the  total  internal  work  of  the  six 
beams,  each  of  which  is  in  a  state  of  flexure,  but  not  under 
thrust.  The  work  of  shear  will  be  neglected.  Evidently'  the 
internal  work  UA,  of  beam  A,  is  double  that  of  one  of  its  halves, 
from  symmetry  (in  this  case).  A  similar  statement  applies  to 
Us',  for  B';  while  for  the  two  beams  B  and  B  the  internal 
work  UB  is  four  times  that  of  the  half  length  of  one  of  them. 
It  remains  to  fill  in  the  detail  of  the  relation 

dU      d 

.      W  "37 


1   C^(dM\  ,  I 

mj  M(wH  -°- 

For  beam  A, 

2  f*. 

--  \      T/V      r/7r 

~EIJQ   '  ~ 


'  dM  dUA      2      *.  27a3 


r  • 


dV  dV  ~EI  ~  3EI' 

For  the  two  beams  B,  B, 

Vx    dM     x         dUB       4       b  Vx    x          763 


M  —    2    5       ^T7    ^O  5     •*• 


For  beam  B' ', 

7'x     /P-27\        dM 
2   ~  \~~2~~r5     d7"=~:c; 


and 


dUB'       2     C 
~d^  =  ETJ0 


b(P-2V)x  2Vb*      Pb* 

~x>dx~ 


dV  ~E'I'JQ  2 

Substituting  these  values  in  eq.  (123)  we  obtain 


80 


MECHANICS    OF    INTERNAL    WORK 


§58 


If  all  the  beams  are  of  the  same  material  and  the  same 
moment  of  inertia  of  cross-section,  and  a  =  b,  this  reduces  to 


Example  I.  Find  the  value  of  V  for  the  case  in  Fig.  30 
in  case  the  weights  of  the  beams  are  not  neglected  (weight  per 
running  foot  being  w  for  beam  A,  and  wf  for  each  of  the  others). 

Example  II.  Instead  of  the  concentrated  load  applied  at 
the  middle  of  beam  A,  apply  a  load  W  distributed  uniformly 
over  the  whole  length  of  beam  A,  in  Fig.  30;  first,  neglecting 
the  weights  of  the  beams;  then,  considering  those  weights. 

For  cases  resembling  that  of  Fig.  30,  but  in  which  the 
arrangement  is  not  symmetrical,  the  mode  of  treatment  will  be 
the  same  but  evidently  more  detail  will  be  involved. 

58.  Trussed-Beam  (or  Brake-Beam).  In  Fig.  31  at 
(A)  is  shown  a  straight,  prismatic,  homogeneous  beam  with 
its  two  extremities  resting  on  two  piers  at  the  same  level; 


(B) 


provided  with  a  short  vertical  strut,  DK,  at  its  middle  point, 
and  also  with  two  oblique  tie-rods  connecting  the  lower  end 
of  the  strut  with  the  two  extremities  of  the  beam.  It  is  to  be 
understood  that  these  four  members  are  accurately  fitted  and 
secured  to  each  other,  before  loading,  in  the  position  shown, 
but  not  so  as  to  create  any  stress  in  any  member;  in  other 
words,  the  beam  is  still  straight,  and  stresses  will  begin  to  be 
produced  in  oil  the  members  as  soon  as  any  load  is  placed 
upon  the  beam;  that  is,  there  are  no  "  initial  stresses." 

In  the  present  case  the  only  load  is  the  concentrated  vertical 
load  P,  at  the  point  Z);  when  bearing  which,  the  structure  is 
assumed  to  be  at  the  same  temperature  as  when  the  parts  were 
put  together.  It  is  required,  by  the  method  of  "  Least  Work," 
to  find:  The  tensile  stress,  T\,  induced  in  each  tie-rod;  the 


§  58  TRUSSP:D-BEAMS  81 

compressive  force  produced  in  the  strut,  DK;  the  thrust,  Tr 
produced  in  the  horizontal  beam  itself;  and4  an  expression  from 
which  we  can  compute  the  "  bending  moment,"  M,  at  any 
section  of  the  beam.  As  regards  the  total  internal  work,  U, 
of  all  the  members,  that  of  the  strut  (on  account  of  its  short 
length)  will  be  neglected,  that  of  each  tie-rod  will  be  an  expres- 

f^  'T7  2 

sion  for  the  case  of  simple  tension,  viz.,  — ^ — ,  while  that  for 

CT2 

the  beam  will  consist  of  a  term  — ~—  for  the  work  of  the  thrust, 

/M2dx 
9y ,  ,  for  the  work  of   the  "  bending  moments  "; 

(that  due  to  the  shears  being  neglected). 

Here  C\  denotes  cri=r  (see  §  20),  where  l\  is  the  length  and 

FI  the  sectional  area  of  each  tie-rod;  E\  being  the  modulus  of 
elasticity.  Also  C  =  l  +  FE,  where  I  (  =  2a)  is  the  length,  and  F 
the  sectional  area,  of  the  horizontal  beam  itself;  and  E  the 
modulus  of  elasticity  of  its  material.  The  angle  DBK  is  denoted 
by  a  (see  Fig.  31). 

At  (B)  in  Fig.  "S&,  is  shown  the  free  body  nOnf  with  all  the 
forces  external  to  it,  we,  have  from  simple  statics  (since  P/2 
is  evidently  the  reaction  of  each  support), 

P 
-  2X  ay 

dM 
(so  that  -Jm-=  —x  •  sin  a);    which  are  to  be  substituted  in  the 

expression  for  internal  work  of  the  whole  structure,  viz. : 


By  differentiating  this  last  expression  with  respect  to  T^ 
we  have 


82  MECHANICS    OF    INTERNAL    WORK  §  58 

which,  after  substitution  of  above  relations,  is  to  be  put  equal 
to  zero,  according  to  eq.  (49).     We  thus  progressively  derive 

2CiTi+CTi  (cos  a)  cos  a 

2_f*-*/Px  .      \ 

+  #/Jo       V2  "  *)(*:.$ 

or, 

2  sin  a  Ca  \P 

Ti[2Ci  +  C  cos2  a]  -  '  VJ     I      -^-x2dx  - TI (sin 
^y    J0    [.2 

that  is, 

<^n«f  _^^  (l 


and  finally,  solving  for  7\,  the  tension  in  either  tie-rod,  noting 
that  a  is  only  the  half-length  of  the  beam,  we  obtain 

T  =  _  Pa3  sin  «  ,     __ 
+3C  cos2  a)  +2o*  sin2  a 


T,  and  the  thrust  in  DK,  are  now  easily  found,  and  the  value 
of  M  at  any  point  of  beam;  and  finally  the  maximum  unit 
stress  (which  should  be  below  the  elastic  limit)  in  the  outer 
fibers  of  the  beam;  as  due  to  combined  thrust  and  bending 
moment. 

If  all  members  of  the  system  are  of  the  same  material  they 
expand,  or  contract,  together,  in  case  the  temperature  changes 
after  the  parts  are  put  together;  but  if  the  beam  were  of  timber, 
for  example,  and  the  tie-rods  iron  or  steel  (as  is  often  the  case), 
the  metal  would  expand  so  much  more  than  the  timber,  for  a 
rise  of  temperature,  that  the  maximum  moment,  and  stresses 
dependent  on  it,  in  the  beam,  might  be  much  larger  than  if 
the  temperature  had  remained  at  the  original  value  (see  next 
paragraph).  Similarly,  a  fall  of  temperature  might  greatly 
alter  the  stresses. 

59.  Cambered  Trussed-Beam  ;  and  Effect  of  Change  in 
Temperature  in  an  Ordinary  Trussed  Beam.  If  it  is  desired 
to  give  a  "  camber  "  or  slight  curvature  (convex  on  upper  side) 
to  the  beam  of  the  foregoing  article,  before  it  is  loaded,  this 
may  be  done  by  making  the  vertical  strut  a  little  longer  than 
is  called  for  by  the  exact  fitting  of  all  the  parts  without  strain. 


59 


TRUSSED  BEAMS 


83 


In  such  a  case,  to  find  the  stresses  in  the  tie-rods,    etc.,  after 
loading,  use  may  be  made  of  §  32,  as  follows : 

In  Fig.  32  the  tie-rods  BK  and  KO  are  shown  in  such  a 
position  that  the  distance  from  B  to  0  is  horizontal  and  equal 
to  the  length  of  the  beam  (between  joint  centers).  The  strut 
KD,  of  a  length  a  little  greater  than  for  no  camber,  being  placed 
vertically  with  lower  extremity  at  K,  its  upper  end  D  projects 
a  small  distance  D'D,  =  A',  above  the  line  joining  B  and  0, 
and  hence  when  the  axis  of  the  beam  is  placed  so  as  to  pass 
through  B  and  K,  with  one  extremity  at  B,  the  other  extremity 
will  be  found  at  0'.  If  now  we  join  KO'  and  let  fall  the  per- 
pendicular 0  .  .  .  e  upon  the  line  KO',  we  (practically)  deter- 
mine the  distance  O'e,  =  ^,  as  the  amount  by  which  the  tie-rod 


FIG.  32. 

on  the  right  is  too  short  to  connect  the  "  joints  "  K  and  0'. 
This  is  called  ^0  in  §  32,  the  bar  in  question  being  a  redundant' 
member.  Hence  the  stress,  TI,  in  the  tie-rod,  after  the  parts 
are  forced  together  and  the  structure  loaded  (at  the  same 
temperature),  may  be  found  from  eq.  (57)  of  §  32;  that  is,  by 
putting  dU/dTi  =  ^0,  the  stress  TI  being  assumed  to  be  tension. 
From  Fig.  32  we  have  O'e=~00'  .  sin  a  =  2Diy  .  sin  a;  i.e., 
^-S^sincj, 

We  may  therefore  find  the  stress  TI  in  the  tie-rod  by  putting 
^  instead  of  zero  in  the  right-hand  member  of  eq.  (126);  thus 
obtaining 


T 


EI(6d  +3C  cos2  a)  +  2a3  sin2  a 


If,  with  the  uncambered  trussed-beam,  the  temperature,  t'  , 
after  loading,  is  lower  than  that,  tQ,  when  the  parts  were  put 


84  MECHANICS    OF    INTERNAL    WORK  §  60 

together,  it  becomes  evident,  from  a  figure  drawn  to  represent 
the  (straight)  axes  of  the  beam,  the  strut,  and  one  of  the  tie- 
rods,  as  assembled  (unstrained)  at  the  new  temperature  t', 
that  the  other  tie-rod  is  too  short  to  fit  into  place  by  some 
amount,  ^o,  T\  may  be  found  from  eq.  (128)  by  insertion 
therein  of  the  value  of  ^0.  For  example,  if  the  coefficient  of 
expansion  or  contraction  for  the  steel  tie-rods  is  >?,  while  that 
of  the  timber  beam  is  negligible  in  comparison,  as  also  that  of 
short  strut,  we  have, 


If  the  temperature  tf  of  loading  is  higher  than  the  original 
temperature  to  of  "  assembling,"  with  steel  tie-rods  and  timber 
beam,  we  replace  the  ^  of  eq.  (128)  by  the  (negative)  quantity  : 
-2y(1f  -to)li;  for  instance,  if  £'  =  35°,  and  fo  =  15°,  C.,  with 
Zi  =  10  ft.  --=120  in.  we  have 

^0=  -2(0.000011)(35  -15)(120)  =  -0.053  inch. 

60.  Other  Cases  of  Trussed-Beams.  Example  I.  Find 
the  stress  in  one  of  the  tie-rods  of  the  trussed-beam  in  Fig.  31 
in  case  the  loading,  instead  of  being  concentrated  in  the  middle 
D,  is  a  uniformly  distributed  load,  W,  covering  the  whole 
length  (2)  of  the  beam  at  rate  w  Ibs.  per  lin.ft.  Disregard 
changes  of  temperature. 

5wa4  sin  a 
l  =24Ci  +  12C  cos2  a)  +8a3  sin2  a 


Example  II.  In  Fig.  31  suppose  the  single  concentrated 
load  P  to  be  placed  at  a  point  in  the  right-hand  half  of  the 
beam  and  at  a  distance  =  c  from  support  0.  Find  T\,  the 
stress  in  tie-rod  OK. 

Example  III.  Trussed-Beam  with  Two  Struts  and  Three 
Tie-rods.  Fig.  33  shows  this  case;  the  load,  W,  =3wa,  being 
uniformly  distributed  over  the  whole  length,  3a,  of  the  beam; 
which  is  prismatic  and  homogeneous  and  is  provided  (at  points 
dividing  its  length  into  thirds)  with  two  short  vertical  struts, 
and  three  tie-rods  1,  2,  and  3,  as  shown;  tie-rod  2  being  hori- 
zontal. The  figure  shows  the  notation  as  to  dimensions  and 
angles.  The  internal  work  of  the  struts  will  be  neglected,  as 
also  the  work  of  shear  in  the  beam.  Rods  1  and  3  are  equal  in 


00 


TRUSSED-BEAMS 


85 


sectional  area  FI  and  length  /i  (  =  aseca),  their  modulus  of 
elasticity  being  E\  ;  corresponding  quantities  for  rod  2  being 
F2,  E2,  and  12  (  =  a);  and  for  the  beam,  F,  E,  and  I  (  =  3a) 
(with  /  as  moment  of  inertia  of  section)  . 


Let 


E 
r  \ 


be  denoted  by 


bY  ^2;  and  ^^  by  C. 


The  four  members,  assumed  to  be  fitted  together  closely, 
without  strain,  before  loading,  form  an  elastic  system  with  one 
redundant  bar.  Bar  (rod)  1  will  be  taken  as  the  redundant 


w=  3  wa 


ill  will-ill  1 1 II 


FIG.  33. 

bar;  and  the  stress  in  it,  T\,  is  to  be  determined,  for  the  given 
loading;  change  of  temperature  being  disregarded.  Hence  by 
§29  we  write  dU/dTi  =  Q.  Let  T^the  thrust  in  the  beam 
(evidently  the  same  at  all  sections),  and  M  the  (variable) 
"  moment  "  in  any  section;  and  T2,  the  tensile  stress  in  rod  2. 
In  the  present  case  we  have,  therefore, 

dU  dT* 


dM 


dM 


From  the  free  body  situated  on  the  left  of  a  vertical 
cutting  plane,  passed  anywhere  between  0  and  H  and  at  a 
distance  x  from  0,  we  find 


and 


wx2 
- 


whence 


dT 


dM 
and     -77^"=  —x  sin  a. 


86 


MECHANICS    OF    INTERNAL    WORK 


§61 


Also,  from  the  free  body  extending  from  0  to  any  vertical 
cutting  plane  passed  between  H  and  K,  x  measured  from  0  as 
before,  we  have 

wx2 
T2  =  T,     and     M  =   wax-        - 


that  is,     T2  = 
whence 


cos  a, 
dT2 


wx* 


and     M  =  %wax  — T\a  sin  a ; 


,     dM 

=  cos  a,     and     -rsfr=  —a  sin  a. 
al  i 


These  various  functions  of  TI  and  corresponding  derivatives 
being  substituted   in   the   detailed   expression   just   given   for 

-J77T,  the  various  integrations  performed,  and  the  result  placed 
al  i 

equal  to  zero;  there  is  obtained,  on  solving  for  T\, 


llwa4  sin  a 


6EI[2Ci  +  (C2  +  C)  cos2  a]  +  10a3  sin2  a 


(129) 


If  the  temperature  under  load  is  different  from  the  original 
temperature,  the  new  value  of  TI  may  be  obtained  in  the 
manner  already  illustrated  in  §  59. 

61.  Straight  Prismatic  Beam  Supported  by  Two  Piers  and 
by  Parabolic  Cable.  (Fig.  34.)  The  extremities  of  the  beam 
rest  on  two  fixed  piers  0  and  C,  situated  at  the  same  level 
(and  capable  of  preventing  upward  as  well  as  downward  move- 


(B) 


ment  of  the  ends  of  the  beam)  and  the  beam  is  attached  by 
(very  numerous)  vertical  suspension  rods,  uniformly  spaced 
along  the  horizontal,  to  a  cable  fastened  at  its  extremities  to 
the  two  fixed  points  E  and  F,  which  are  at  a  common  level 


§  61  BEAM  SUSPENDED  FROM  CABLE  87 

and  vertically  over  0  and  C.  It  is  stipulated  that  before  any 
load  acts  on  the  beam  (its  own  weight  and  that  of  rods  and 
cable  being  neglected)  all  the  suspension  rods  are  barely  taut, 
as  also  the  cable  (from  one  point  of  attachment  of  a  rod  to 
the  next),  and  that  the  curve  formed  by  the  cable  is  a  common 
parabola,  with  its  geometrical  axis  vertical  and  passing  midway 
between  points  E  and  F.  R  is  the  vertex  of  the  curve.  That 
is,  the  beam  and  the  suspension  rods  and  the  cable  are  closely 
fitted  to  each  other  initially,  and  without  strain,  so  that  if 
a  vertical  load  is  gradually  applied  to  the  beam  the  stresses 
in  the  beam  and  rods  and  the  parts  of  the  cable  increase  pro- 
gressively, and  proportionally,  from  zero  up  to  their  final  values. 
Evidently  this  presents  a  statically  indeterminate  problem; 
involving  but  one  redundant  element,  if  we  consider  the  bending 
of  the  beam  under  load  so  slight  that  the  curve  of  the  cable 
remains  practically  a  parabola.  For  this  character  of  curve 
it  is  known  from  mechanics  that  the  tensions  in  the  suspension 
rods  are  all  equal  if  these  rods  are  very  numerous  and  uniformly 
spaced  along  the  horizontal,  as  in  the  present  case.  The  deforma- 
tion (stretching)  of  the  cable  and  suspension  rods,  when  the 
beam  is  loaded,  will  here  be  neglected,  as  having  but  small 
influence  compared  with  the  bending-  of  the  beam. 

Let  now  a  uniformly  distributed  load,  W ',  of  rate  w'  per 
foot  run,  be  placed  on  the  beam,  extending  from  0  along  a 
portion  OD,  =a,  of  the  length,  Z,  of  the  beam;  i.e.,  W '  =  w'a;  w' 
and  a  being  given.  This  will  produce  reactions,  viz.,  V  at  0 
and  V  at  (7,  from  the  piers,  and  equal  stresses  in  all  the 
vertical  rods  at  same  rate  w  per  horizontal  foot.  The  system 
of  forces  now  acting  on  the  beam,  as  "  free  body/'  is  shown 
at  (E)  in  Fig.  34.  There  are  three  unknowns,  viz.,  V,  V ',  and 
w}  while  statics  provides  only  two  independent  equations; 
hence  the  third  must  depend  on  elastic  theory.  As  the  single 
"  redundant  element  "  we  may  take  either  V,  V,  or  w\  let 
us  choose  w;  then  the  others  must  be  expressed  as  functions 
of  w.  From  the  free  body  at  (B)  (whole  beam),  by  moments 
about  C  we  find 

wl 

•  •  •  •  •  <130> 


88  MECHANICS    OF    INTERNAL    WORK  §  61 

If  the  vertical  rods  are  b  ft.  apart,  the  stress  in  some  par- 
ticular rod,  which  is  to  be  taken  as  the  redundant  bar,  is  ivb] 

and  by  the  principle  of  least  work  (§  29)  we  may  put    ,   ^  =0: 

dU 

i.e.,  7—  =  0. 
;  dw 

The  internal  work  of  shear  being  neglected,  and  also  that  of 
the  cables  and  vertical  rods  (as  previously  implied),  we  have, 
therefore  (E  and  7  constant  for  beam), 


--    • 

dw  JD          dw 

For  the  dx  at  any  point  n  on  portion  OD  of  beam,  x  being 
measured  from  0,  taking  On  as  a  free  body,  we  have  [see  also 
eq.  (130)], 


[w'a(     a\     wl~\ 
TV  2/~tj 


wx2    w'x2 
"~2 2~' 


dM         I       x2       x2-lx 
and  hence      ~~[w~  =  ~2X+~2'  =   ~~2  —  '      '     '     *    '"-    *     '     ' 

while  for  any  point  n'  on  portion  DB,  with  x  measured  from  0, 
and  On'  as  free  body, 


.     ...     (134) 

[w'a  /_     a\     twZ"l       wx2  I      a\ 

-rV-2J-yJx+—  -^-2)5 

dM         I       x2       x^-lx 

and      •'•  ar'-ff+t-  =  ^~"  '  '  '  I  iS  (  ^ 

Hence  substitution  in  eq.  (131)  gives 

w'x2]  /x2  —  lx 


Ca  r  ^'a2.r     it?Zx     i^a:2     w'x 

J,  r"'ijr  -~+— 


w'a?x     wlx     wx2 
-2T  -2-+-^ 


§  61  BJ  AM  SUSPENDED  FROM  CABLE  89 

.     Performing  the  integrations  between  respective  limits  indi- 
cated, and  reducing,  we  find 


-  2w'a5  +  5t//a4Z  -  5w/a2Z3  +  2wl5  =  0  ; 
solving  which  for  w,  writing  m  for  the  ratio  a+l,  we  have 

fy)2 

w  =  ^-(4m3  -  5m2  +  10)w'.  (136) 

With  w  known,  V  and  V,  and  the  moment  and  shear  at  any 
.section  of  the  beam,  are  easily  computed,  in  a  numerical  case. 

[Note.  Evidently  this  problem  involves  the  case  of  one 
variety  of  suspension  bridge;  in  a  more  thorough  treatment 
of  which,  however,  the  deformation  of  the  cable  should  not  be 
neglected;  see  Professor  Hiroi's  book  on  "  Statically  Indeter- 
minate Structures  "  for  a  fuller  investigation.  The  relation  in 
eq.  (136)  is  confirmed  by  a  statement  in  Mr.  H.  M.  Martin's 
small  work  of  same  title  as  Professor  Hiroi's;  and  the  results 
given  in  the  remainder  of  this  paragraph  are  quoted  from  Mr. 
Martin's  book.] 

If  a  single  concentrated  load  P  is  carried  on  the  beam  in 
Fig.  34,  instead  of  a  distributed  load,  being  placed  at  any 
distance  c  from  the  left  support  0,  similar  analysis  (writing  n 
for  the  ratio  c-^-Z)  leads  to  the  result 


;    .....     (137) 

while  the  reaction  V  at  the  support  C  is  found  to  be 

7/  =  tn(10n2-5n3-3)P.     ....     (138) 

For  example,  if  c  =  JZ,  V  =  — &P  (points  downward). 

If  the  beam  is  built  in  horizontally  both  at  0  and  0 
("  fixed  ends  ")  instead  of  merely  resting  on  the  piers  ("  hinged 
ends  "),  with  original  close  fit  without  strain  as  before,  the 
same  treatment  may  be  adopted  but  the  detail  is  much  more 
cumbersome.  We  have  now  three  redundant  elements;  viz., 
w  and  the  moments  M0  and  Mc  of  the  two  internal  couples 
in  the  beam  at  0  and  C.  If  any  one  of  these,  say  M0,  be 


90  MECHANICS    OF    INTERNAL    WORK  §  62 

regarded  as  a  single  redundant  element,  the  stresses  in  the 
rods  and  the  forces  forming  the  other  couple  may  be  treated 
as  independent  loads,  like  P  itself  (see  §  53),  justifying  us  in 

placing  -        =0.     Similarly,  we  have   ,,,   =0,  and  -,  —  =  0;  and 

ttiVi  o  CLIVL  c  (*W 

thus  derive  three  independent  equations,  aside  from  the  rela- 
tions  of  statics.  In  this  way  w,  M0,  and  Mc  can  be  found. 

If  in  this  case  (  "fixed  ends  ")  the  load  is  uniformly  dis- 
tributed, viz.,  W'  =  w'a,  extending  a  ft.  from  support  0,  we  find 
(with  m  =  a-r-i), 

'\  .     :     .     .     (139) 


and  the  shear  in  the  vertical  section  of  the   beam   close  to 
support  C  is 

Jc^m3(7m  -4  -3m2)w'l.      .     .  _.     .     (140) 

Also  the  value  of  Mc  in  section  at  C  is 

.     .     .     .     .     (141) 


(A  positive  result  indicates  tension  in  upper  fibers.) 

Again,  if  the  beam  in  Fig.  34  (A)  has  fixed  ends  and  carries 
a  single  concentrated  load,  P,  at  distance  c  from  left  end  O 
by  similar  process  we  find  (denoting  c  +  l  by  n), 

P 

,       ....     (142) 


Jc  =  n2(28n-12-15n2)P,    ....     (143) 
and  Mc=Jn2(3-8tt  +  5^2)PZ.      .,.     .     .     (144) 


62.  Double-Knee  Beam  ;  under  Side  Pressure.  Constant 
E  and  I.  Fig.  35  shows  this  "  double-knee  beam  "  as  it  may 
be  called.  It  is  a  continuous  beam,  homogeneous,  and  of 
.constant  moment  of  inertia  of  section  (7),  consisting  of  three 
straight  portions,  continuous  at  trie  "  knees,"  G  and  D:  two 
vertical  and  one  horizontal.  Its  own  weight  is  to  be  neglected. 
Its  extremities  are  hinged  to  two  immovable  piers,  at  same 
level,  0  and  B\  and  before  any  load  or  force  is  brought  upon 
it,  it  is  under  no  strain]  that  is,  it  does  not  have  to  be  "sprung" 
to  fit  it  over  the  hinge  pins.  (It  must  be  noted  that  there 
are  no  hinges  at  the  corners  G  and  D.) 


§62 


''DOUBLE-KNEE"  BEAM 


91 


Let  now  a  horizontal  wind  pressure,  wh,  act  against  the- 
left-hand  side,  of  intensity  of  w  Ibs.  per  linear  (vertical)  foot. 
It  is  required  to  determine  the  horizontal  component  H,  of  the 
hinge  reaction  at  0,  the  vertical  component  at  0  being  V 
(components  at  B  are  H'  and  V)  by  the  foregoing  principles. 
Notation  as  in  figure.  Also  let  F  =  common  sectional  area  of 

all  parts  of  beam  and  let  C\  denote  -7^77;  and  C%  denote  7™ 

Denote  by  T\,  Tz,  and  T3,  the  (unknown)  thrusts  in  por- 
tions OG,  GD,  arid  DB,  respectively;  and  neglect  the  internal 
work  due  to  shears. 

This  is  a  statically  indeterminate  elastic  structure:  with 
only  one  redundant  element,  however,  and  this  may  be  taken 
as  either  H,  or  Hf\  say,  H. 

We  shall  therefore  put  "TJ^O; 

which  may  be  justified  in  either 

of   two   ways.      Note    that    of 

the   five   external  forces,  wh  is 

fixed,  i.e.,   given;    and   that  V 

and   V  are  also  fixed   and  are 

not    affected    by  any  imagined 

change  in  H  or  H'\    since  by 

taking    moments    about    B  we 

have  V  =  wh2  +  2l.     If  the  sup-  FIG.  35. 

port  at  0  consisted  of  a  fixed 

horizontal  smooth  surface,  a  change  in  H  would  simply  mean 

a  shifting  of  joint  0  of  beam  along  this  surface  and  the  beam 

would  still  be  supported;    H  may  therefore  be  considered  as  a 

load,  independent  of  P,  and  must  have  a  value  corresponding 

to  the  fact  that  in  the  present  case  joint  0  of  the  beam  does 

not   move  horizontally  during  the   gradual  application   of  the 

loads  wh  and  H  (H'  would  simply  be  a  "  neutral "  reaction 

during  this  application;  see  §  21);  i.e.,  dU  +  dH  =  Q,  by  §  21. 

Again,  from  another  point  of  view,  consider  that  the  beam 
is  connected  with  the  fixed  support  on  the  left,  not  by  a  hinge 
joint,  but  by  two  inelastic  bars,  one  horizontal,  the  other  ver- 
tical; then  the  stress  in  the  former  would  be  H]  in  the  latter, 
V.  These  bars  must  be  inelastic,  otherwise  joint  0  of  the  beam 


92  MECHANICS    OF    INTERNAL    WORK  §  63 

would  move  during  the  gradual  application  of  the  load  wh\ 
and  are  now  to  be  considered  as  part  of  the  structure.  Since 
the  beam  is  only  hinged  to  the  support  at  B,  and  not  "  built-in/' 
it  is  evident  that  the  vertical  bar  is  a  necessary  bar  and  not 
redundant,  since  without  it  the  beam  would  not  be  supported. 
But  the  horizontal  bar  is  a  redundant  bar;  since,  if  it  were 
omitted,  the  beam  would  still  be  supported,  the  joint  0  simply 
shifting  to  a  position  a  little  to  the  right  of  the  original.  Hence 
by  §  29  we  may  put  dU  +  dH  =  Q;  in  which  U  should  include 
the  internal  work  of  the  two  bars;  but  these  being  inelastic, 
such  internal  work  is  zero,  so  that  in  this  case  U  simply  refers 
to  the  internal  work  of  the  beam  alone.  We  shall  therefore 
use  the  relation  dU  +  dH  =  Q  and  solve  for  H;  remembering  that 
V  does  not  depend  on  H  and  hence  need  not  be  replaced  by  its 
value  in  terms  of  wh,  until  toward  the  close  of  the  algebraic 
work. 

From  §  37,  eq.  (74),  we  have 


and  hence 

dU  dK  dT*  dT»      1   r"         dM 


^  ClT 


dT»      1   r 
3dH+EiJ0 


For   the   ds's   situated    along   OG,    ds  =  dy;     T\  —  V\  and 

=  Hy-^~.     For  those  on  GD,  ds  =  dx-,    T2=wh-H;  and 

wh2 

=  Hh  —  ^  —  Vx;    while   along  DB,   ds  =  dz;     T3  =  V;  and 


We  have,  therefore, 

fl>  -         (  +  y)dy  = 


CD* 
Jo 


§  62  "  DOUBLE  -KNEE  "    BEAM  93 


Hh*     wh*     Vlh2 


12        2    ' 

Noting  also  that  both  dTi+dH  and  dT^^-dH  are  zero, 
while  dT2+dH=  —  1,  and  substituting  in  the  expression 
dU^dH  =  0,  we  find 


2         * 


A  similar  problem  is  treated  by  Mr.  Mensch,  on  p.  80  of 
Engineering  News  of  Feb.,  1900.  Besides  the  wind  pressure 
on  the  vertical  side  Mr.  Mensch  also  considers  a  uniformly 
distributed  load  along  the  whole  length  of  the  horizontal  por- 
tion GD,  of  intensity  of  w\  Ibs.  per  linear  foot.  He  not  only 
neglects  the  internal  work  due  to  shears  but  also  that  due  to 
the  thrusts.  In  his  problem  the  moment  of  inertia  of  the 
section  of  the  portion  GD  is  different,  viz.,  /i,  from  that,  /, 
of  the  vertical  portions  of  beam.  His  final  result  for  H  is, 
writing  n  for  the  ratio  I\  -=-  /, 


.. 


The  results  in  eqs.  (145)  and  (146)  are  seen  to  agree  when  in 
(145)  we  make  C2  =  0  (implying  that  the  effect  of  the  thrusts 
is  negligible  in  this  connection),  and  in  (146)  place  w\  equal 
to  zero  and  I\  equal  to  /. 

If  the  horizontal  side  pressure  is  P  Ibs.  concentrated  at  the 
corner  G,  instead  of  the  distributed  pressure  wh,  it  may  easily 
be  proved  that  H  =  H'  =  $P. 

Example.  If,  in  the  case  of  Fig.  35,  the  loading  consists 
of  one  vertical  load,  P,  applied  vertically  at  the  middle  of  GD, 
find  the  value  of  H]  and  also  the  vertical  displacement  of  the 
point  of  application  of  P, 


MECHANICS   OF    INTERNAL    WORK 


63 


63.  Loaded  Davit  of  Uniform  Section.  Displacement  of 
Extremity.  The  davit  [see  (A)  in  Fig.  36]  consists  of  a  straight 
vertical  portion  DK,  built  in  vertically  in  the  fixed  support  at 
D,  and  the  curved  portion  KO,  which  is  the  quadrant  of  a  circle 
of  radius  r.  E  and  7  are  assumed  constant  throughout  the 
whole  length  DKO.  The  tangent  at  K  is  vertical  (before  load- 
ing). Neglecting  the  weight  of  the  davit  itself,  it  is  required 
to  find  the  vertical  displacement  Oa,  =?yi,  of  the  extremity  0, 
as  due  to  the  gradual  application  of  the  vertical  load  PI. 
Before  loading,  the  extremity  is  at  0;  after  loading,  at  0'. 


~ 


n/\\   af  T° 

r    XPS       •  .,1 

d     xN     ^PI 

P  X    \     I  /S<1.. 


FIG.  36. 

Since  the  projection  of  the  displacement  00'  upon  the  line 
of  the  external  force  or  load  PI  is  the  quantity  sought,  viz., 
Oa,  =2/1,  it  may  be  determined  as  the  derivative  of  the  total 
internal  work,  U,  of  the  beam,  with  respect  to  PI;  i.e.,  by 

€q.  (25)  of  §  21,  or  yi  =jp~-     Neglecting  the  work  of  shear,  and 

also  in,  this  case  the  work  of  thrust,  we  have  only  the  internal 
work  due  to  bending  moments,  Af;  and  hence  may  write 
[from  eqs.  (74)  and  (75),  §  371, 


dM 


M  - 


dM 


(147) 


For  the  ds's  situated  along  DK  we  measure  z  from  D;    so 
that  ds  =  dz.     For  those  along  the  curved  part  KO,  using  polar 


§  63  LOADED    DAVIT  95 

.coordinates,  as  shown,  with  angle  0  measured  from  radius  CK, 
we  have  ds  —  r  .  dd. 

For  any  point  nf  on  DK  we  have 

.  .    .  •• 

M  =  Pir:     -nT-  =  r:     and     ds  =  dz\ 


while  for  any  point  n  on  the  curve  KO  we  have,  from  the  free 
body  nO,  shown  at  (B)  in  Fig.  36, 

M  =  Pir'cosd:    -775-  =  r  cos  6  ;     and     ds  =  rd6. 


Substitution  in  eq.   (147)  gives  (see  integral  forms  in  the 
Appendix), 


_0  -  •  (147a) 

To  find  the  horizontal  displacement  of  0,  viz.,  aO'  in  (A), 
Fig.  36,  since  there  is  no  external  force  at  0  pointing  hori- 
.zontally  we  have  simply  to  introduce  an  imaginary  force  PI 
at  0  directed  horizontally  toward  the  right,  and  use  the  rela- 

tion aO'  =  2/2  =  -7D~,  m  which  finally  P%  is  to  be  made  equal  to 

OIL  2 

:zero;    thus  following  the  procedure  of   §§  23a  and  40.     With 
P2  as  well  as  PI  in  action  we  have 

For  any  point  n'  on  DK  : 


For  any  point  n  on  quadrant  KO  : 

M  =  Pir  cos  d  +  P2r[l  -sin  0];     (—r  =  r(l  -sin 

(i/    2 


MECHANICS    OF    INTERNAL    WORK 


§64 


Substituting  in  y2  =  dM/dP2   (making,  however,  P2  =  0  in 
each  of  the  expressions  for  M,  before  substitution),  we  derive 
p  r  r  rh  /v/2  n 

y2  =  -L.\  J    (h+r-z)<b+i*l       (cos  0 -sin  0  cos 


(/.  +  r-r.V.     ,    .     (148). 


64.  Semicircular  Curved  Beam,  or  Arch  Rib  ;  Hinged  at 
Ends.  Single  Eccentric  Load.  The  curved  beam  is  of  semi- 
circular form,  radius  =  r,  and  is  assumed  to  be  homogeneous 
and  of  constant  moment  of  inertia.  That  is,  E  and  /  are  con- 


FIG.  37. 

stant.  Its  own  weight  is  neglected.  It  is  hinged  at  its  extrem- 
ities to  the  fixed  supports  0  and  K,  at  the  same  level,  being 
fitted  to  the  hinges  without  strain,  at  temperature  to,  before  load- 
ing. The  temperature  remaining  constant,  a  single  eccentric 
vertical  load  P  is  then  placed  at  the  point  m,  60°  from  tho 
right-hand  horizontal  radius,  CK.  Evidently  the  line  of  P  is  f  r 
from  0  and  jr  from  K.  It  is  required  to  find  the  horizontal 
component,  H,  of  the  reaction  at  each  hinge  (evidently,  in  this 
case,  H  is  the  same  for  both). 

Considering  the  whole  rib  as  a  free  body,  we  easily  find  by 
ordinary  statics  (by  moments  about  0  and  K)  that  the  vertical 
components  of  the  reactions  at  0  and  K  are  }P  and  fP, 
respectively  (they  are  independent  of  H)]  that  is,  H  con- 
stitutes the  only  redundant  element  in  the  problem;  and  may 

be  found  by  putting  -777  =  0  (bringing  into  play  the  same  method 
of  proof  that  was  used  in  the  fore  part  of  §  62). 


§  64  SEMI-CIRCULAR   ARCH    RIB.      END    HINGES  97 

Here  the  work  of  the  thrust,  T,  will  be  considered,  as  well 
as  that  of  the  "  bending  moment,"  M,  but  the  work  of  shear 
will  be  neglected.  With  F  as  the  sectional  area  of  the  beam, 
we  have  from  eqs.  (74)  and  75), 


™       j 
Mdffd8 


For  any  point  n  between  0  and  m,  with  6  as  variable  angle 
(see  Fig.  37),  we  have,  from  the  free  body  On  (like  that  at 
(B)  in  Fig.  38), 

dM 

M  =  \Pr(\  -cos  0)  -Hr  sin  0;      --=  -r  sin  6\ 


dT 

~      =  *[n  6]    ds==rd6'> 


6 


while  for  any  point  n'  on  part  Km,  reckoning  the  corresponding 
variable  angle  6'  from  radius  CK,  n'K  being  the  free  body, 


dT 

=       cos 


/=fPsinfl'-#cos  0'. 
Hence,  with  ds  =  rdd,  or  rdO',  we  find 


sn  +    sn  cos 


dU     r3   ri20° 

dH=in  j     (~*p  si 

r3     /^60° 

+  E~1  1      (  ~ 

r     rim* 

+  —  —  I        (  \P  sin  OcosO  +  H  sin2 
r  /i^o 


sin  P  +      sin  ^  cos  ^+£r  sin2 


sn      cos 


98  MECHANICS    OF    INTERNAL    WORK  §  65 

Performing  the  integrations  (see  Appendix),  collecting  terms 
and  reducing,  and  placing  dU/dH  =  Q,  we  find 

0.  .     (149) 

Writing  I  =  Fk2,  in  which  k  is  the  radius  of  gyration  of  the 
cross-section  of  the  beam,  and  solving  for  H  there  results 

•    '•     (150) 


H  being  found,  the  thrust,  shear,  and  bending  moment  are 
easily  determined  for  any  section  of  beam. 

The  radial  thickness  of  the  beam  being  small  compared 
with  the  radius  r,  the  effect  of  considering  the  work  of  thrust 
in  this  semicircular  form  is  extremely  slight.  Disregarding  the 
work  of  thrust,  which  simply  means  dropping  the  term  affected 
by  the  factor  (r+FE)  in  eq.  (149),  or  (which  is  the  same  thing) 
neglecting  the  term  k2  in  (150)  compared  with  r2,  we  have 

#  =  0.2386P,      .     .     .     .     .     .'    (151) 

instead  of  (150).  For  example,  with  A:  =  5  in.,  and  r  =  20  ft.  = 
240  in.,  the  two  results  differ  by  about  one-tenth  of  one  per 
cent,  only.  Even  when  the  form  of  rib  is  a  fairly  flat  circular 
arc,  the  effect  of  the  thrust  terms  is  relatively  small. 

The  effects  of  a  change  of  temperature  will  be  treated  in 
§66. 

65.  Semicircular  Arch  Rib  with  Tie-Rod.  In  this  case 
two  fixed  piers  with  smooth  horizontal  surfaces  furnish  vertical 
reactions,  only,  at  the  ends  of  the  semicircular  rib  (constant  E 
and  I  as  before)  whose  extremities  are  connected  by  a  horizontal 
bar  or  tie-rod.  Piers  at  same  level.  (See  (A),  Fig.  38.) 

When  the  load  P,  to  be  placed  at  the  "  crown/'  m,  or  highest 
point  of  the  rib,  is  applied,  the  ends  0  and  K  of  the  rib  spread 
apart  slightly,  and  at  the  same  time  the  horizontal  tie-rod  is 
stretched;  the  final  stress  in  the  latter  being  7\,  which  is  to 
be  determined;  it  being  understood  that  the  rib  and  tie-rod 
are  fitted  closely  together  without  strain,  at  some  temperature 
to,  before  the  load  is  applied,  and  that  the  temperature  does 


65 


SEMI-CIRCULAR    ARCH    RIB    WITH    TIE-ROD 


99 


not  change.  Let  ^i  denote  the  sectional  area,  li  the  length, 
=  2r,  and  EI  the  modulus  of  elasticity,  of  the  tie-rod;  C\  repre- 
senting li~FiEi. 

Evidently,  in  this  elastic  system,  the  tie-rod  is  a  redundant 
bar;   there  being  but  one  redundant  element  in  the  case;   and 

hence  we  write  j^r^O?  by  §  29.     The  reactions  (vertical)  of 

the  two  smooth  piers  are  each  P/2,  from  symmetry,  and  inde- 
pendent of  TI.  Considering  the  work  of  thrusts  (including  TI) 


d   f  CK  M2ds      CK  T^dsl 
dflljo    ~2ET+Jo    2FE\  = 
2    Cm    dT 

TTTI  I         •*•  jrji  ^*"        '-'•  • 

tijo      dl  i 


FIG.  38. 

and  of  moments,  but  not  that  of  shear,  we  have  from  eqs.  (74) 
and  (75),  etc., 

dU       o 


(152) 


(On  account  of  symmetry,  we  are  enabled  to  put  the  internal 
work  of  the  rib  equal  to  twice  that  of  the  half  between  0  and  m.) 

To  find  moments  and  thrusts  in  terms  of  TI  consider  the 
portion  of  the  beam  from  0  to  any  point  n  in  the  left-hand  half, 
as  a  free  body;  see  (B)  in  Fig.  38.  M  is  found  by  taking 
moments  about  the  gravity  axis  of  the  section  at  n\  and  T7, 
the  thrust,  by  summing  components  parallel  to  the  tangent 
/.-..£;  thus  excluding  the  shear  J  from  either  sum. 

We  have,  then,  for  points  on  the  quadrant  Om, 

dM 
M  =  vVPr(l-cos  0) -TV  sin  0;  ^r=  -rsin  0; 


4^—sin0;     ds  =  rdO. 


100  MECHANICS    OF    INTERNAL   WORK  §  66 

Substitution  in  eq.  (152)  gives 

9r3  /"«/2 

(71y1  +       I       ( _  jp  sin  e  +  JP  sin  0  cos  0  +  TI  sin2  6>)d0 
^Vo 

2r  A/2 
+-^  J       ( JP  sin  0  cos  0  +  T7!  sin  20)d0  =  0.      .     (153) 

Performing  the  integrations  (see  Appendix)  and  reducing, 

^   P     siT.       rP       «r2V 
~£/  •  2  +  2  EI+2FE+2FE    u' 


(154) 


66.  Temperature   Stresses   for   Semicircular   Arch   Rib. 

If  the  rib  is  hinged  to  two  fixed  piers,  as  in  Fig.  37,  and  the 
temperature  t  under  load  is  higher  than  that  fo  of  "  erection  " 
(when  the  rib  was  fitted  in  place  without  strain),  we  may  con- 
sider the  piers  to  furnish  only  the  vertical  reactions,  and  that 
the  horizontal  reactions  are  due  to  an  inelastic  horizontal  rod 
or  bar  connecting  0  and  K,  whose  coefficient  of  expansion  is 
zero,  and  which  was  fitted  in  place  without  strain  at  the  original 
lower  temperature  tQ.  Then  if  the  two  parts,  rib  and  bar, 
were  to  be  fitted  together  at  the  higher  temperature  it  would 
be  found  that  the  bar  is  too  short  by  Ao  =  2rr)(t—  to),  y  being  the 
coefficient  of  expansion  of  the  material  of  the  rib;  since  under 
no  constraint  the  distance  between  the  extremities  of  the  rib 
has  increased  by  that  amount. 

Hence  by  eq.  (57)  we  may  put  dU/dH=Ao,  and  in  this,  U 
must  include  the  internal  work  of  the  bar;  but  as  the  latter 
is  zero  U  comprises  the  internal  work  of  the  rib  alone.  There- 
fore, for  the  loading  of  §  64  and  Fig.  37  we  have  only  to  put 
2rj)(t—  to)  in  place  of  the  zero  in  the  right-hand  member  of 
eq.  (149)  and  solve  for  //;  obtaining  (in  case  the  work  of  both 
thrust  and  shear  in  the  rib  is  disregarded), 


.     .     .     (155) 


§  66  SEMICIRCULAR    ARCH    RIB.      TEMP.    STRESSES  101 

If  to  is  lower  than  t,  the  same  formula  holds  but  t—  to  is 
negative.  If  the  rib  in  Fig.  37  carries  no  load,  the  H  is  entirely 
due  to  a  change  of  temperature,  if  any;  having  a  value 


In  Fig.  38,  where  horizontal  constraint  is  exerted  upon  the 
semicircular  rib  by  an  actual  bar  or  tie-rod  OK,  which  is  elastic 
and  has  a  coefficient  of  expansion  T?I  (different,  it  may  be,  from 
that,  )?,  of  the  material  of  the  rib),  it  would  be  found  that  in 
fitting  rib  and  rod  together  at  a  temperature  t,  higher  than 
that,  to,  of  erection,  the  rod  is  too  short  by  an  amount 
io*"2r00  —  ifc)  (t  —to)  to  connect  the  extremities  of  the  rib.  Hence 
we  put  dU/dTi  =  fa,  U  including  the  internal  work  of  the 
tie-rod.  In  other  words,  we  write  2r(r}  —  rn)(t-to}  in  the  place 
of  the  zero  in  the  right-hand  member  of  eq.  (152)  ;  and  obtain 


(156) 


In  case  the  rib  in  Fig.  38  is  not  loaded  this  reduces  to 

•      (157) 


For  a,  fall  of  temperature  (t—  to)  would  be  negative,  while  if 
i)i  were  numerically  greater  than  y,  (y  —  yi)  would  be  negative. 

67.  Semicircular  Curved  Beam  without  Hinges  ;  E  and  I 
Constant.  Single  Eccentric  Load.  [See  (A)  in  Fig.  39.] 
This  means  that  the  semicircular  arch  rib  is  "  built  in,"  or  inserted 
rigidly  in  the  two  immovable  piers  (at  same  level)  at  0  and  K 
without  strain  before  loading,  and  is  continuous  between. 
No  change  of  temperature  considered  at  present.  Sectional 
area  of  rib  is  F.  A  vertical  load  P  is  to  act  on  the  rib  at  point  m, 
60°  from  the  horizontal  radius  CK  on  the  right.  Hence,  when 
the  load  is  on,  the  cross-sectional  plane  of  the  rib  at  0  (also 
at  K)  will  be  the  same  as  before  loading,  and  the  center  of  gravity 
of  the  section  at  0  (and  at  K)  will  not  move  as  the  load  is  applied. 

At  0,  therefore,  when  the  load  is  on,  there  will  be  in  action 
a  couple,  of  moment  M0;  a  horizontal  shear,  H;  a  vertical 


102 


MECHANICS    OF    INTERNAL   WORK 


67 


thrust,  Vj  all  unknown  in  advance.  Similarly,  in  the  section 
at  K,  Mk,  V,  H']  also  unknown.  Since,  for  the  whole  beam 
as  a  free  body,  statics  furnishes  only  three  independent  equa- 
tions (one  of  which  gives  immediately  H'  =  H],  and  there  are 
six  unknowns,  it  is  evident  that  the  beam  is  statically  inde- 
terminate to  the  extent  of  three  redundant  elements;  and  let 
V,  H,  and  M0  be  chosen  as  these  three  redundant  elements. 

If  the  rib  has  been  duly  fitted  in  place  without  strain,  and 
then  the  load  P  applied,  it  is  evident  that,  the  beam  being 
built  in  at  K,  the  pier  at  0  might  be  entirely  removed  and  the 
beam  would  still  be  supported,  though  becoming  deformed  (it 
would  be  a  cantilever).  The  forces  of  the  couple,  also  V  and  H, 
might  then  be  gradually  applied  as  independent  loads  with  the 


f  I    H 

M0t        (B) 


r        c 


FIG.  39. 


result  that  the  sectional  plane  at  0  and  the  center  of  gravity 
of  that  section  would  finally  be  brought  into  the  position  where 
the  pier  0  constrains  them  to  be  under  actual  conditions  under 
load.  In  other  words,  we  might  remove  pier  0,  let  the  rib 
remain  unloaded  at  first,  thus  leaving  the  part  0  of  beam  in 
the  same  position  as  it  will  be  finally  under  actual  load,  and 
then  gradually  and  simultaneously  apply  all  these  "  loads,"  viz., 
P,  V,  H,  and  the  forces  of  the  couple,  until  they  reach  their 
respective  final  values.  During  this  gradual  action  the 
points  of  application  of  V  and  H  would  not  be  displaced,  and  the 
cross-sectional  plane  of  the  beam  at  0  would  not  turn,  thus  justi- 
fying us  in  putting,  separately  [see  eqs.  (25)  and  (34)], 


-0      =0  and 


(158) 


§  67  SEMICIRCULAR   ARCH   RIB.      NO   HINGES  103 

It  is  to  be  noted  that  these  are  three  independent  equations. 
For  instance,  it  is  not  necessary,  before  performing  the  differen- 
tiation implied  in  dU/dH  =  Q,  to  express  the  other  quantities 
MQ,  and  V,  in  terms  of  H\  since  the  three  quantities,  M0,  V, 
and  H,  are  independent  of  each  other.  For  example,  we  might 
assume  arbitrary  values  for  MQ  and  V,  and  then  the  imposing 
of  the  condition  dU/dH  =  0  would  simply  give  us  such  a  value 
of  H  that  during  the  gradual  loading  just  described  the  point  0 
of  the  beam  would  not  be  displaced  horizontally;  although  it 
would  be  displaced  vertically,  and  the  section  would  turn 
through  some  angle  <j>  ;  as  dependent  respectively  on  the  arbi- 
trary choice  of  V  and  M0.  Hence  if  there  is  to  be  neither  ver- 
tical nor  horizontal  displacement  at  0,  nor  any  turning  of  the 
plane  of  the  cross-section,  all  three  of  the  equations  (158)  must 
hold,  independently.  It  remains  to  supply  the  detail  of  apply- 
ing the  three  equations  in  (158),  in  order  to  determine  the 
three  unknown  constants,  MO,  V,  and  H. 

Neglecting  the  internal  work  of  shear,  but  considering  that 
due  to  thrust  T  and  moment  M,  we  have 


Evidently  the  rib  must  be  divided  into  two  parts,  0  .  .  m 
and  m  .  .  .  Ky  for  the  summations  required.  For  any  point  n 
on  0  ...  m,  taking  the  free  body  shown  at  (B)  in  Fig.  38,  we 
have 

M  =  M0  +  Vr(l  -cos  6)  -Hr  sin  6] 
and  hence 

dM  dM  dM 

T5rr  =  l;    -yf7-  =  r(l  -cos  0);     and     -jvp=—  rsm0. 

dMQ  dV  dH 


Also  T  =  Fc 

dT  dT  dT 

whence  *    ~dV  =  cos    '    a         ~dff==sm 


104  MECHANICS   OF   INTERNAL   WORK  §  67 

For  any  point  n'  between  m  and  K,  with  0  .  .  .  n'  as  free 
body,  still  reckoning  angle  0  from  radius  CO  (noting  that  the 
distance  of  P  from  0  is  fr;  from  K,  %r),  we  have 

M  =  M0  +  Vr(l-cos0)-Hrsin  0  +  Pr(J+cos  0); 
hence 

dM  dM  dM  . 


T=V  cos  0  +  #  sin  0  -P  cos  0 
and  therefore 

7  dT7  dT 

0  =  cos0      and          =  sm 


First  substituting  in  eq.  (159),  noting  that  the  second  inte- 
gral disappears  since  dT/dMo  =  Q,  we  have,  with  ds  =  rd6, 

r    ri2o° 
-^7  (M0  +  Vr  -  Vr  cos  0  -Hr  sin  0)dO 

Mi L   /n° 
•x  u 

r    r i8o° 
+1T7  I        (M0  +  Fr  -  7r  cos  0  -Ifr  sin  0  +  £P +Pr  cos  0)d0  =  0. 

Performing  the  integrations  (see  Appendix),  noting  that 
sin  120°  =  0.8660,  cos  120°=  -0.500,  arc  120°  =  §TT,  and 
arc  180°  =  TT,  we  obtain  after  final  reduction, 


0;    .     .     (162) 


which  is  the  first  of  the  necessary  three  simultaneous  equations 
involving  the  three  unknowns  M0,  H,  and  V. 
Next  we  have  from  eq.  (160), 


Substitution  in  this  equation  of  the  values  for  M,  dM/dV, 
dT/dV  already  obtained,  having  regard  to  the  portion  of  beam 
within  which  each  applies,  with  ds  =  rdO,  leads,  after  integration 
reduction,  to  the  result, 

=0.     (163) 


§  67  SEMICIRCULAR    ARCH    RIB.      NO    HINGES  105 

Similarly,  the  result  of  substituting  in  eq.  (161),  or  dU/dH 
=  0,  i.e.,  in 


.after  integration  and  reduction  is  found  to  be 


The  three  simultaneous  eqs.  (162),  (163),  and  (164),  can  now 
be  solved  for  M0,  V,  and  H. 

Work  of  Thrust  Neglected.  With  such  a  large  angle  as  180° 
for  a  circular  arch  rib  the  effect  of  considering  the  work  of 
thrust  is  extremely  small;  so  the  terms  arising  from  that 
source  will  now  be  neglected  (that  is,  the  terms  involving  the 
divisor  F).  This  causes  I  to  disappear  from  all  three  equa- 
tions. By  elimination  between  (162)  and  (163)  we  find 

7  =  0.1955P  (Ibs.);         .    ,     .     .     .     (165) 
and  further  combination  gives 

ff  =  0.312P(lbs.);      .     ,    I     ..    .     .     (166) 
;and 

M0  =  0.1127Pr  (in.4bs.)  .....     (167) 

Work  of  Thrust  Retained.  If  the  terms  containing  the 
divisor  F  be  retained  and  eqs.  (162),  (163),  and  (164)  be  com- 
bined without  modification  (that  is,  the  work  of  thrust  is  now 
•considered;  though  not  that  of  shear)  the  result  of  elimination 
will  be 


0.307, 

(168) 


(0.0466-  Y~2 

and  H=- ^j— ; (169) 

0.1491+?.-^ 


106  MECHANICS    OF    INTERNAL    WORK  §  68 

and,  further,  regarding  both  V  and  H  as  having  been  found, 


If  in  eqs.  (168)  and  (169)  we  write  I  =  Fk2,  where  k  is  the 
"  radius  of  gyration  "  of  the  cross-section  of  rib,  the  fraction 
I+Fr2  reduces  to  k2/r2,  which  ratio  is  extremely  small;  and 
the  terms  containing  it  are  so  small,  relatively  (in  this  case 
of  a  semicircular  rib  with  radial  thickness  small  compared  with 
the  radius  r  of  the  semicircle)  that  the  omission  of  the  terms 
involving  this  fraction,  I  +  Fr2,  in  the  equations  mentioned, 
leads  to  no  practical  error  and  gives  rise  to  the  results  already 
obtained  in  eqs.  (165),  (166),  and  (167).  For  results  in  case 
the  rib  is  segments!  instead  of  semicircular,  and  for  tem- 
perature stresses,  see  §§68  and  69. 

68.  Segmental  Circular  Arch  Rib  Hinged  at  Ends.  Constant 
E  and  I.  Fig.  40  shows  this  case,  with  a  single  concentrated 
load  P  at  any  point  m,  the  radius  to  which,  Cm,  makes  an 
angle  /?  with  the  horizontal  radius  CO  on  the  left  (ft  may  be 
<  or  >90°).  As  before,  the  rib  is  homogeneous  (constant  E) 


Angle 


and  its  cross-section  has  a  constant  7;  the  radial  thickness 
being  small  compared  with  the  span  OK.  The  radius,  Cn  of 
any  point  n  of  the  axis  of  rib  on  the  left  oftm  is  at  some  angle  0 
with  the  horizontal  measured  from  CL;  while  for  any  point  n' 
on  the  right  the  corresponding  angle  is  called  0',  measured  from 
CS.  The  angle  SCm  will  be  denoted  by  /?';  i.e.,  p^x-p. 
The  axis  of  the  rib  is  the  arc  of  a  circle  of  radius  r,  and  is 
symmetrical  about  the  vertical  line  through  center  C.  With 
the  one  vertical  load,  P,  as  shown,  the  reaction  of  each  hinge 


§  68  HINGED    SEGMENTAL   CIRCULAR    ARCH    RIB  107 

is  oblique  (since  the  rib  tends  to  spread),  and  will  be  replaced 
by  its  vertical  and  horizontal  components.  From  S  (hor. 
compon.)=0  the  two  horizontal  components  are  equal  to  each 
other  (value  H).  The  vertical  reactions  PQ  and  PK  are  easily 
found  by  statics.  With  the  whole  rib  as  free  body,  moments 
about  hinge  K  gives,  after  canceling  the  r, 

(COS0Q  +  COS/?)P 

Po=         2  cos  00 

The  piers  are  immovable  and  it  is  assumed  that  the  rib 
has  been  fitted  in  place  originally  without  strain  and  the  load 
applied  later;  and  that  no  change  in  temperature  has  occurred. 
The  value  of  H  is  the  statically  redundant  element  (there  is 
but  one  such  element  in  this  problem)  and  may  be  found,  as 
already  explained  in  §  62,  by  putting  dU/dH  =  Q.  As  before, 
let  ds,  =rd6  (or  rdO',  on  right  of  P),  bo  an  elementary  portion 
of  the  axis  of  the  rib,  at  any  point  n  (or  nr)  ;  then,  considering 
the  internal  work  of  bending,  and  of  thrust,  but  not  that  due 
to  shear,  we  have  [see  eq.  (75)], 


-«*+     •*-*  •  <172> 


the  summation  being  extended  over  the  whole  length  of  the 
axis  of  rib,  from  0  to  K.     Detail  as  follows: 

For  any  ds  occurring  between  0  and  m  we  have,  consider- 
ing portion  On  as  a  free  body, 

M  =  P0r(cos  00  -cos  0)  -Hr(sm  0  -sin  00),     .     (173) 

and  -777--=  —r  (sin  0—  sin  0o),      ....    (173a) 

art 

also  T  =  P0  cos  0  +  H  sin  0,     .     .     .     .     (174) 

dT 
and  '=:  +sm  ^'  ...   (174a) 


and  the  limits  for  0  are  from  0  =  OQ  to  0  =  /?. 

Similar  expressions  for  M  and  T  will  hold  for  any  ds  between 
m  and  K,  if  for  P0,  0,  and  dd,  we  write  Pk,  0',  and  dO',  respect- 
ively; and  the  limits  for  0'  will  be  0o  and  /?'. 


108  MECHANICS    OF    INTERNAL    WORK  §  68 

Substitution  being  made  in  eq.  (172)  we  obtain 

t 

tcos  #o  sin  6  -sin  6  cos  0  -sin  00  cos  00  +sin  00  cos  6]dO 


-&T  I    [sin2  ^  -2  sin  00  sin  6  +  sin2  00]d0 
^ 


i  sin  ^  cos  e  do  +JE  (  sin2  ede 

[cos  00  sin  0'  -  sin  6'  cos  0'  -  sin  00  cos  00  +sm00cos0']d0' 


rrr  I     [sin2  0'  -2  sin  00  sin  ^  +sin2  dQ]dO' 


^in  ^  cos  d'de'  + 


To  effect  a  solution  of  eq.  (175)  for  H,  we  first  perform  the 
integrations  indicated;  which  is  fairly  simple,  since  each  inte- 
gral contains  but  one  variable,  6  or  6',  and  the  forms  may  be 
found  in  the  table  of  integrals  in  the  Appendix.  For  PK, 
P—Po  is  now  substituted;  and  then,  for  PQ  (in  each  term 
containing  it),  the  value  given  in  eq.  (171).  Terms  are  now 
collected;  and  use  made,  for  further  reduction,  of  the  following 
relations:  /?'  =  TT-/?;  sin  ^  =  sin  0;  cos  /?'=  -cos  /?; 

sin  2/?'=  -sin  2/?  and  I  =  Fk?. 

If  now  we  denote  by  A  the  expression 

J(sin2  /9  -3  sin2  00)  +sin  00[sin  8  -  0i  cos  00  +/?i  cos  /?], 

li 
result, 


in  which  Oi=--6Q  and  ^=-—  p,  final  solution  leads  to  the 


(A;2 
A-^.-^[sm2t3-sm2 

H  =  -  -^-  -.   .     (176) 

0i  (1  +2  sin2  00)  -|  sin  200  +  ^  (00  +  sin  200) 

With  H  known,  the  values  of  moment,  thrust,  and  shear  are 
easily  found  for  any  section  of  the  rib. 


§  69  HINGED    SEGMENTAL    CIRCULAR    ARCH    RIB  109 

If  the  internal  work  of  thrust  is  disregarded,  the  result 
for  H  is  that  due  to  the  omission  of  the  terms  in  (176)  contain- 
ing the  factor  k2/r2;  i.e., 

/-t  tjn\ 

H==dl(l+2  sin2  60)  -I  sin  200' 

As  the  arc  of  the  arch  rib  approaches  a  semicircle  the 
difference  between  (176)  and  (177)  becomes  very  slight. 

69.  Temperature  Stresses  in  Circular  Segmental  Arch  Rib  of 
Two  Hinges.  With  load  P  still  in  place  (in  Fig.  40),  the  length 
of  span  being  OK  =  2r  cos  OQ,  if  the  temperature  rises  to  a  value 
t  higher  than  that,  fo,  of  erection,  the  piers  being  unyielding, 
we  may  prove,  as  in  the  first  part  of  §  66  that  the  value  of  H 
is  obtained  by  putting  dU/dH  equal  to  the  amount  ^0,  viz., 
y(2r  cos  #o)(£— fo),  that  the  distance  between  the  hinges  0  and 
K,  of  the  rib,  would  have  increased,  had  the  rib  been  free  to 
expand  with  rising  temperature  (instead  of  putting  dU/dH  =  Q). 
Filling  in  details  in  the  same  manner  as  in  §  68  we  obtain  (for 
meaning  of  A,  see  end  of  §  68), 


-i  4>inV-sin20o])P  +  -^(2)?cos  60)(t-to) 

H=- — ~ ^ ;     (178) 


#i(l+2sin2  00)-f  sin  2#0-f-^(#i+sin 

j^ 

in  which  if  P  be  made  zero  the  value  of  H  is  that  resulting 
from  the  change  of  temperature  alone.  Of  course  in  that  case 
PO,  PK>  and  -P  (m  Fig.  40)  would  drop  out;  and  the  moment, 
thrust,  and  shear  at  any  section,  due  to  the  H  alone,  would  be 
the  elements  from  which  the  fiber  stresses,  which  now  may  be 
called  "  temperature  stresses,"  would  be  computed. 

For  a  fall  of  temperature  below  that,  to,  of  erection,  the 
quantity  (t—  to)  becomes  negative  in  eq.  (178). 

70.  Segmental  Circular  Arch  Rib  without  Hinges.  E  and  I 
Constant.  Ends  Fixed  in  Immovable  Piers.  Here,  as  in  §  67, 
it  is  stipulated  that  the  ends  of  the  rib  are  fixed,  or  "  built  in  " 
(or  "  encastre"  ")  in  the  piers  or  supports  without  strain  before 
any  loading;  that  is,  the  rib  is  put  in  place  originally  without 
initial  stresses.  Fig.  41  shows  this  case,  a  symmetrical  arrange- 


110 


MECHANICS    OF    INTERNAL    WORK 


§70 


ment  with  extremities  0  and  K  at  the  same  level;  E  and  I 
constant.  A  single  vertical  load  P  is  placed  at  any  point  ra, 
the  radius  to  which,  Cm,  makes  an  angle  /?  with  the  left-hand 
horizontal  radius  LC.  Each  of  the  extreme  radii  CO  and  CK 
makes  an  angle  00  with  the  horizontal;  and  the  angle  LCK 
(which  =  K  -  OQ)  will  be  called  //. 

The  variable  angle  OCn  for  any  point  n,  or  element  ds,  of 
the  axis  of  rib  on  the  left  of  m  will  be  called  0.  For  any  point 
n',  on  the  right  of  m,  also,  the  angle  0  will  be  measured  from 
LC  in  the  present  analysis. 

When  the  load  P  is  in  place  (weight  of  rib  itself  neglected) 
stresses  will  be  induced  in  the  cross-section  at  0  equivalent  to 


FIG.  41. 

a  couple  of  moment  M0,  a  shear,  and  a  thrust.  But  the  shear 
and  thrust  may  be  replaced  by  a  single  resultant  passing 
through  the  center  of  gravity  of  the  section  at  0,  and  this 
resultant  may  then  be  decomposed  into  a  horizontal  component 
H  and  a  vertical  component  T7.  These  are  shown  in  Fig.  41  as 
acting  on  that  end  of  the  rib.  Similarly,  in  the  cross-section 
at  K,  the  stresses  acting  are  equivalent  to  a  couple  of  moment 
MK,  a  horizontal  force  HK  and  a  vertical  force  VK',  these  latter 
acting  through  the  center  of  gravity  of  the  cross-section  at  K. 
These  two  couples,  the  forces  H,  HK,  V,  VK,  and  the  load  P, 
form  a  system  of  forces  holding  the  whole  rib  in  equilibrium 
when  considered  as  a  free  body.  It  is  seen  that  (P  being 
given)  six  unknowns  are  involved,  viz.,  M0,  MK,  H,  HK,  V, 
and  VK-  Since  statics  furnishes  but  three  independent  equa- 
tions, the  remaining  three  needed  for  complete  solution  must  be 
based  on  the  theory  of  elasticity.  As  in  §  67  we  note  that  there 


§  70  SEGMENTAL    CIRCULAR    ARCH    RIB.      NO    HINGES  111 

are  here  three  redundant  elements  involved,  which  could  be 
any  three  of  the  above  six  quantities.  Our  present  treatment 
will  select  MO,  H,  and  F,  as  the  three  redundant  elements;  and 
the  analysis  will  be  so  shaped  as  to  exclude  the  quantities  MR, 
HK,  and  VK  from  the  equations  brought  into  play  for  deter- 
mining M0,  H,  and  V.  We  shall,  therefore  (see  §  67),  write 
the  three  relations 


and  from  them  determine  MO,  F,  and  H,  in  terms  of  P. 

Details  as  follows  :    From  the  free  body  0  .  .  .  n,  we  have 
for  the  moment  and  thrust  at  any  section  n  between  0  and  m, 

M  =  Mo  +  Fr(cos  60  -cos  d)  -Hr($m  6  -sin  00)  ;       -     (180) 
T=Fcos0+#sin0;     .     .     ........     (181) 

whence 

dM  dM  dM 

=  1  ;    -      =  r  (cos  00  -cos  0)  ;    -r    =  -r  (sin  0  -sm  00)  ; 


.and 

dT  dT  dT 


whereas,  for  any  section  n'  between  m  and  K  we  have,  from 
O  ...  n'  as  a  free  body, 

M  =  MO  +  Fr(cos  00  —cos  0)  —  #r(sin  0  —sin  00)  —  Pr(cos  /?  —cos  0) ; 

-whence 

dM  dM  dM 

-jiif  =  1;    -Tf7r=r(cos  0o— cos  0);    ~jrr  =  — r(sm  0— sin  0o); 
divi  o  dv  tin 

-and 

dT  dT  dT 


112  MECHANICS    OF    INTERNAL   WORK  §  70' 

First  Equation.  Expanding  the  first  statement  of  eq.  (179) , 
neglecting  the  work  of  shear,  and  indicating  limits  of  itegra- 
tion  by  limiting  values  of  the  variable  angle  6  for  the  two  por- 
tions Om  and  mK  of  the  rib,  we  have 

dU       I    C^-dM  1    r?dT  , 


1    C'\,dM      i        1    f 
-— I    M         .  ds+__| 
tilj*      dM0  FEJdo 


K&MA      J_fV^  -n  nQtt 

*dMo+F£j,  2dM0dS 

But,  from  eqs.  (182)  and  (184),  dT/dMQ  is  zero  both  on 
0  .  .  .  m  and  on  m  ...  K:  hence  the  second  and  fourth  integrals 
on  the  right  drop  out.  Substituting,  now,  the  proper  values 
of  M  and  dM/dM0,  from  eqs.  (180)  to  (184),  in  the  first  and 
second  integrals,  putting  ds  =  r  .  dd,  and  denoting  the  expression 
forming  the  right-hand  member  of  eq.  (180)  by  Z,  we  have, 
after  canceling  the  r/E7, 


0.  (186) 

rz  r*  C* 

Note.     Now    I   Zdd+  I   Zdd  is  the  same  thing  as    I   ZdO; 

^do  J ft  Jo<> 

whence  we  have 

0-cos  6)  -Hr(sm  d  -sin  dQ)]dO 

-Pr  P(cos.^-cos  0)dO  =  Q.          (187) 
*$ 

This  equation  contains  but  one  variable,  0.  After  inte- 
gration, with  insertion  of  limits,  further  reduction  is  brought 
about  by  the  use  of  the  relations:  /JL  =  TT  —  #o;  sin  /*  =  sin  #o; 

cos  /£==  -cos  OQ\   we  shall  also  write  ^  —00=01,  and  ^—  /?=/?i. 

The  final  form  then  becomes 

20i  (M0  +  Vr  cos  00)  -2#r[cos  00  -6l  sin  60] 

-Pr[(^i+/?i)  cos^+sin/?-sin  00)  =  0;     .     (188) 

which  is  the  first  of  the  required  three  equations  containing 
the  three  unknown  constants  MO,  V,  and  H  (and  no  others). 
The  given  quantities  are  P,  0Q,  61,  /?,  fa,  E,  7,  F,  and  r. 


§  70  SEGMENTAL    CIRCULAR    ARCH    RIB.      NO    HINGES  113 

Second  Equation.     Expanding  dU/dV  =  0  [of  eqs.  (179)]  we 
have  (neglecting  work  of  thrust) 

dU__ 
dV~l 


Values  for  M,  T,  dM/dV,  and  dT/dV,  from  eqs.  (180)  to 
(184),  are  now  inserted  in  their  proper  places  in  (189);  and 
advantage  is  taken  of  the  relation  pointed  out  in  the  "  Note  " 
following  eq.  (186).  Whence,  with  ds  =  rd6,  we  obtain 


r 
I 


-ji  I    [M0  +  Vr  (cos  00  -cos  0)  -Hr(sm  0  -sin  #0)](cos  60  -cos  d)dO 

r3    fn 

—pl  I    P(cos  P  ~cos  ^)(COS  ^o  -cos  0)dO 


+jfi\  J    (^sin^+Fcos^)cos^^-  J    Pcos20dO\=0.     (190) 

Integration  between  the  limits  shown,  and  the  aid  of  rela- 
tions already  mentioned,  reduce  this  to  the  form  (Fk2  having 
replaced  /), 

2/^tMo  cos  #o  -2#r(cos2  00  -Ol  sin  00  cos  60) 

[(      /A  T 

20  1  cos2  #0  +  (^1  +j3/(?i  -i  sin  2^o)J 

-Pr    ffi  +/?i)  cos  ,5  -sin  60  +  sin  ^]  cos  00  ~  (sin  00  -sin  /?)  cos 

i+/?i  -J-(sin  20o  +  sin  2^9)]  =  0;  (191) 


which  is  the  second  equation  needed;   between  Af0,  F,  and  H. 

Third  Equation.     Finally,   the   detail   of   dU/dH  =  0   [see 
eqs.  (179)]  is  as  follows  (work  of  shear  omitted)  : 

dU      1   r>       dM         1   f"       dM 

=  M  •  -ds  +  M  '  dffds 


114    .  MECHANICS    OF    INTERNAL    WORK  §  70 

Substituting  values  from  eqs.  (180)  to  (184)  and  utilizing 
the  relation  of  the  "  Note  "  following  eq.  (186),  with  ds  =  r  .  dO, 
we  have 

r2     fit 

•wj  I   [M0  +  Fr(cos  00  -cos  6)  -Hr(sm  d  -sin  00)](sin  00  -sin  0)d0 

&*JOo 

rs  rft 

—  F7  1    ^(cos  p  -cos  0)(sin  00  -sin  0)d0 
tiijB 

+•£$1    M#sin0  +  Fcos0)sin0d0-  J    Pcos0sin0d0    =0. 

After  integration  and  reduction  this  becomes  * 
2A/0(0o  sin  00  -cos  00)  +2Fr  cos  00(0i  sin  00  -cos  00) 

fl  +:j)('i  +1  sin  200)  -2  sin  200-f  20i  sin2  00 

sin  00cos/?+sin  00sin/?-cos/9  (cos  00  +  c 

0.     ...     (194) 


(193) 


This  is  the  third  equation  needed,  of  the  three  between  the 
unknown  constants  MO,  V,  and  H. 

Elimination.  By  a  fortunate  coincidence  eqs.  (188)  and 
(191)  happen  to  be  so  constituted  that  both  MO  and  H  may  be 
eliminated  by  a  single  operation,  viz.  :  Multiply  eq.  (188) 
throughout  by  —cos  00  and  add  the  resulting  equation  to 
(191),  member  to  member.  The  equation  thus  produced  con- 
tains neither  MO  nor  H\  and  solved  for'F  gives  rise  to 


cos  p  (sin  p  -sin  00)  +  1  1  +—  )#  \P 


in  which  the  symbol  B  stands  for  the  expression  * 

J(0i +0i -Jam  200 -ism  20).      .     .     .     (196) 

Similarly,  M0  and  V  may  be  eliminated  from  the  two  equa- 
tions (188)  and  (194)  by  one  operation,  viz. :  Solve  each  of  the 

*  For  meaning  of  0t  and  ^  see  after  eq.  (187). 


§  70  SEGMENT  AL   CIRCULAR    ARCH    RIB.      NO   HINGES  115 

equations  (188)  and  (194)  for  M0  and  equate  the  expressions 
so  derived.  The  resulting  equation  contains  H,  but  not  MQ 
nor  V.  Solving  for  H,  we  obtain,  using  the  symbol  G  to  denote 
the  expression 

cos  00(/?i  cos  /?  +  sin  /?)  -Ol  cos2  /?-  J  sin  200), 


(197) 


1  +       (0i  +  1  sin  200)0i  -2  cos 


Since  #  and  F  are  now  known  in  terms  of  P  a  solution  of 
eq.  (188)  for  MO  will  give  the  latter  quantity  in  known  terms; 
thus 


00)F 
+  ([0i  +/5i]  cos  /3  +  sin  /?  -sin  #o)^-]r.      .     (198) 

Great  complication  would  result  from  the  insertion  in  (188) 
of  the  algebraic  values  of  H  and  V  from  eqs.  (195)  and  (197), 
and  no  useful  purpose  would  be  served  thereby.  When  H  and 
V  have  been  determined  numerically  in  any  given  case,  placing 
their  values  in  (198)  leads  directly  to  the  numerical  value  of  MQ. 

With  M0,  V,  and  H  known,  the  moment,  thrust,  and  shear 
(M,  T,  and  /)  may  be  found  at  any  cross-section  of  the  rib 
(including  that  at  K,  where  we  find  MK,  VK,  and  HK)  by 
the  conditions  of  equilibrium  of  a  free  body  extending  from  0 
to  the  section  in  question  (evidently  HK  =  H).  Hence  all 
stresses  are  now  determinate,  as  to  a  single  vertical  load  P 
applied  at  any  point  m  of  the  rib,  Fig.  41.  It  is  understood 
that  the  temperature  when  P  is  in  place  is  the  same  as  that, 
fo,  of  erection. 

If  in  the  foregoing  expressions  for  MO,  V  ,  and  H,  the  ratio 

k2 

-j,  wherever  occurring,  is  placed  equal  to  zero,  the  results  are 

those  that  would  be  obtained  if,  besides  neglecting  the  work  of 
shear,  we  had  also  neglected  the  work  of  thrust.  As  the  form 
of  the  rib  approaches  a  semicircle,  the  work  of  thrust  is  of  less 
and  less  consequence. 


116  MECHANICS    OF    INTERNAL    WORK  §  71 

71.  Temperature  Stresses  in  Segmental  Circular  Arch  Rib 
of  No  Hinges.  Fixed  Supports.  E  and  I  Constant.  (See  Fig, 
41.)  With  no  load  whatever  on  the  rib,  to  being  the  temperature 
of  placing  in  position  ("  erection"),  there  being  then  no  con- 
straint, that  is,  no  "  initial  strains,"  stresses  are  induced  if  the 
temperature  rises  to  some  higher  figure  £;  due  to  attempted 
expansion  on  the  part  of  the  rib.  At  each  of  the  sections  O 
and  Kj  a  couple,  a  thrust,  and  a  shear,  are  developed.  The 
thrust  and  shear  being  combined  into  a  resultant,  and  that 
resultant  decomposed  along  the  vertical  and  horizontal,  evi- 
dently, from  the  equilibrium  of  the  whole  rib  (from  the  con- 
ditions of  statics)  each  vertical  component  is  zero  and  the  two 
horizontal  components  are  equal  (  =  H)  while  the  moments  of 
the  two  couples  are  equal  (  =  M0).  There  are,  therefore,  two 
redundant  elements,  MQ  and  //. 

If  T)  is  the  coefficient  of  expansion  of  the  material  of  the 
rib,  the  point  0  of  the  rib  would  have  moved  horizontally 
toward  the  left  a  distance,  ^,  equal  to  r)(t  —  to)  2r  cos  00  (see 
Fig.  41)  if  the  rib  had  been  free  to  change  form  by  the  removal 
of  the  left-hand  pier  (the  other  pier  remaining  fixed).  Also, 
by  the  fixity  of  the  left-hand  pier,  the  turning  of  the  plane  of 
the  cross-section  of  rib  at  0,  which  would  have  occurred  if  there 
had  been  a  fixed  hinge  joint  at  0,  is  prevented.  Hence  by  reason- 
ing similar  to  that  on  p.  102  (see  also  §  66)  we  must  have  the  con- 
ditions :* 

=0;  and     -**  •—'•- 


which  will  serve  to  determine  the  two  unknowns,  MO  and  H. 

Detail  as  follows  :  From  the  free  body  0  .  .  .  n,  n  being  any 
point  of  axis  of  rib  between  0  and  K,  we  find 


0-sm#0)     -     ....     (200) 
and  T  =  #sin0.     .     .     .     .     .     .     .     .    ,     (201) 

•''  ST0  =  1;    ^=-Ksin0-sin00);      -     -     (202) 

dT  dT 

-nr  =  0;    and    -r^sin  0  .....     (203) 
dH 

*  See  Note  D  in  Appendix. 


§  71  SEGMENTAL    CIRCULAR    ARCH    RIB.      NO    HINGES  117 

In  making  each  of  the  summations  called  for  in  eqs.  (199) 
advantage  can  be  taken  of  the  symmetry  of  form  of  the  rib, 

by  integrating  between   0  =  00   (at  0)   and   6  =  -   (at  highest 

point,  or  "  crown,"  of  the  rib)  and  doubling  the  result;  as  a 
simple  means  of  obtaining  the  summation  for  all  the  ds's  of 
the  rib  axis. 

Expanding  dU/dMo  =  Q,  we  have  (neglecting  work  of  shear), 


*/2 

(204) 


But  dT/dMo  =  0  and  the  second  integral  drops  out.     Sub- 
stituting from  eqs.  (200),  etc.,  we  derive 


-sn 


i.e.,         M0(jj  -00)  -ff/|cos  00  -  (|-fc)  sin  00J  =0,     .     (205) 

or,  putting  ^  -0o  =  0i, 

0iM0  -#r(cos  0o-0i  sin  00)  =0.     .    ..     .     (206) 

Again,  expanding  dU/dH=XQ,   =rj(t—  to)  2r  cos  00,  we  find, 
putting  ds  =  rdQ,  and  neglecting  work  of  shear, 


J1    C*/2 

2bl 

i.e., 

i  r*/2 

^y  I        [M0  -#r(sin  0  -sin  00)][  -r(sin  0  -sin  00)]rd0 

i>l0,    .     .     (207) 


whence,  after  integration  and  reduction,  putting  -~—  0o  = 
[0i  sin  00  -cos  00]M0  +  [0i  sin2  00-sin  200]Fr 


.     .     (208) 


118  MECHANICS  OF    INTERNAL   WORK 

Elimination  of  M0  from  (206)  and  (208)  gives 


-J3-.  y(t-to)eos 


H  = 


With  H  known,  M0  is  found  from  eq.  (206),  i.e., 


§72 


.     (209) 


(210) 


If  the  work  of  thrust  were  neglected  (besides  that  of  shear) 
the  resulting  value  of  H  would  be  obtained  from  eq.  (209)  by 
writing  zero  in  place  of  the  ratio  k2/r2  in  the  denominator  (only)  . 

72.  Simpson's  Rule  for  Approximate  Integration.  In 
applying  the  methods  of  internal  work  it  frequently  happens 
that  summations  must  be  made  which  on  a  strict  mathematical 
basis  are  either  impossible  or  extremely  cumbersome.  In  such 


cases  recourse  may  be  had  to  Simpson's  Rule,  or  other  method 
of  approximate  integration,  as  will  be  illustrated  in  the  next 
paragraph  (§  73).  The  present  paragraph  will  consist  of  a 
reproduction  of  the  statement  and  proof  of  Simpson's  Rule 
taken  from  the  author's  Notes  and  Examples  in  "  Mechanics  " 
(p.  13);  as  follows: 

If  ABCDEFG,  Fig.  42,  is  a  smooth  curve  and  ordinates  be 
drawn  from  its  extremities  A  and  G  to  the  axis  X,  an  approxi- 
mation to  the  value  of  the  area  so  inclosed,  A...D...G...N 
.  .  .  0  ...  A,  between  the  curve  and  the  axis  X,  is  obtained  by 
Simpson's  Rule,  now  to  be  demonstrated.  Divide  the  base  ON 
into  an  even  number,  n,  of  equal  parts,  each  =  Jx  (so  that 


§  72  SIMPSON'S  RULE  119 

ON  =  n  .  Ax) ,  and  draw  an  ordinate  from  each  point  of  division 
to  the  curve,  the  lengths  of  these  ordinates  being  UQ,  u\,  u2,  etc. ; 
see  figure. 

Consider  the  strips  of  area  so  formed  in  consecutive  pairs; 
for  example,  CDE&'C"  is  the  second  pair  in  this  figure 
(counting  from  left  to  right).  Conceive  a  parabola,  with  its 
axis  vertical,  to  be  passed  through  the  points  C,  D,  and  E.  It 
will  coincide  with  the  real  curve  between  C  and  E  much  more 
closely  than  would  the  straight  chords  CD  and  DE;  and  the 
segment  CDE,  considered  as  the  segment  of  this  parabola,  has 
an  area  equal  to  two-thirds  of  that  of  the  circumscribing  paral- 
lelogram CC'E'E.  Hence,  since  the  area  of  this  pair  of  strips  = 
trapezoid  CEE"C"  +  parabolic  segment  CDE,  we  may  put 

Area  of  pair  of  \ 

•       rt-nn       \    =  2 Ax\$(Uz  +  ^4)  +§(^3  — i[^2  +^4j)j; 

strips  Lilt 
which  reduces  to          ...  JJa:[^2  +  4w3+w4]. 

Treating  all  the  ~-  pairs  of  strips  in  a  similar  manner,  we  have 
finally,  after  writing  Ax=  (xn  —XQ)  +n, 

Whole  area  AG" 

(approx.)         [        3n 

The  approximation  is  closer  the  more  numerous  the  strips  and 
the  more  accurate  the  measurement  of  the  ordinates  UQ,  Ui, 
u2,  etc. 

If  the  subdivision  on  the  axis  X  were  "  infinitely  small/'  an 
exact  value  for  the  area  would  be  expressed  by  the  calculus  form 

f^»    
udx.    Hen.ce  for  any  integral  of  this  form,  I          udx,  if  we 
=  XO                                                                                                                                                                     Jx  —  XQ 

are  only  able  to  determine  the  particular  values  (UQ,  u\,  etc.) 
of  the  variable  u  corresponding  respectively  to  the  abscissae  XQ, 
XQ  +  AX,  XQ  +  2Jx,  etc.  (where  Jx=(xn—xo)+n,  n  being  an 
even  number),  we  can  obtain  an  approximate  value  of  the 
integral  or  summation  by  writing 

xn  Xn-Xo 


I 


~ 

Un].   .       .       ,        (211) 


120 


MECHANICS    OF    INTERNAL    WORK 


§73 


As  to  the  meaning  of  n,  note  that  the  first  ordinate  on  the 
left  is  not  u\,  but  UQ\  also  that  while  there  are  n  strips,  the 
number  of  points  of  division  is  n  +  1,  counting  the  extremities  0 
and  N. 

73.  Davit  of  Variable  Cross-section.  Circular  Quad- 
rant. Deflection  of  Extremity.  Use  of  Simpson's  Rule. 
In  Fig.  43  is  shown  a  davit,  0  ...  6,  of  wrought  iron;  its  axis 
being  a  quarter  circle.  All  sections  circular,  but  of  varying 
radii;  smaller  toward  the  free  extremity  (point  6).  Radius  of 
circular  axis  is  r,  =  0  ...  C,  =  50  in. 


O....b=l 


75° 


At  0  the  extremity  is  "  built  in,"  with  tangent  vertical,  in 
a  fixed  support.  At  the  other  end,  6,  a  vertical  load  P,  =  150 
Ibs.,  is  applied;  under  whose  action  the  extremity  6  (of  the 
axis  of  davit)  is  displaced  from  6  to  6'.  It  is  required  to  deter- 
mine the  vertical  projection  6  ...  a  of  this  displacement,  neg- 
lecting the  weight  of  the  davit  itself;  the  value  of  E  being 
28,000,000  Ibs./inA  Let  us  take  seven  different  points  0,  1,  2, 
etc.,  equally  spaced  along  the  axis  from  0  to  6.  The  angles  (0) 
of  the  radii  drawn  to  these  points  from  center  C  are  respect- 
ively 0°,  15°,  30°,  45°,  60°,  75°,  and  90°;  from  the  horizontal 
0 .  . .  C.  The  radii  of  the  cross-sections  (circular)  at  these 
points  are  respectively  TQ  =  1.0  in. ;  r\  =  0.98 ;  r^  =  0.92 ;  r3  =  0.86 ; 
r4  =  0.75;  r5  =  0.62;  and  r6  =  0.50  in.;  while  the  corresponding 
moments  of  inertia  are  /o  =  i^04  =  0.7854  in.4;  /i=  0.7251  in.4; 


§  73  DEFLECTION    OF    DAVIT.      SIMPSON'S    RULE  121 

72  =  0.5692;  /3-0.4298;  /4  =  0.2487;  75  =  0.1162:  and  76  = 
0.0491  in.4. 

As  will  be  seen  later,  values  of  cos2  0  will  also  be  needed  for 
the  seven  angles:  00  =  0°,  01  =  15°,  02  =  30°,  etc.,  up  to  06  =  90°. 
These  values  of  cos2  0  are  respectively,  1.0,  0.9330,  0.7500, 
0.5,  0.25,  0.0670,  and  0. 

Since  the  vertical  distance  6  ...  a  is  the  projection  upon 
the  line  of  the  external  force  of  the  displacement  6  ...  6'  we 
may  obtain  its  value  by  Castigliano's  Theorem  [§  21]  by  writing 

jr77"'  =y'  =W'    '•  •'  :>   •  -'    (212) 

1  f&  M  <LM 

i.e.  [see  §  36],  y  =  E)0   T'dFds;    '     ''•  '     '     '     (213) 

(neglecting  the  internal  work  both  of  thrust  and  shear,  the  radial 
thickness  of  the  curved  beam  being  small  compared  with  the 
radius  of  the  circular  axis). 

From  the  free  body  m  .  .  .  6,  a  radial  section  having  been 
made  at  m,  we  find  by  moments  about  m, 

M  =  Pr  cos  0  ;    and    -p-  =  r  cos  0.      .     .     (214) 
These  being  substituted  in  eq.  (213),  with  ds  =  rdd,  we  obtain 

.....  (215) 


The  integral  contains  two  variables,  0  and  /.  If  I  were  a 
given  algebraic  function  of  0,  this  expression  could  be  sub- 
stituted and  the  integration  performed  (provided  only  known 
integral  forms  were  encountered).  But  in  the  present  case, 
although  the  curved  beam  has  a  smooth  taper  from  0  to  the 
end  6,  and  all  sections  are  circles,  /  is  not  a  known  algebraic 
function  of  0.  However,  the  value  of  /  may  be  computed  from 
the  measured  radius  of  section  for  any  assigned  value  of  0. 
Simpson's  Rule,  therefore,  offers  a  means  of  securing  a  result 
accurate  enough  for  practical  purposes. 


122  MECHANICS    OF    INTERNAL    WORK  §  73 

Comparing  the  integral  in  eq.  (215)  with  that  in  the  left- 
hand  member  of   (211),  we  note  that  dd  corresponds  to  dx', 

n  .    /cos2  6\ 

^  to  xn',   0  to  XQ]    and  I  —  j  —  )  to  u.     Choosing  a  value  ol  o 

for  n,  we  therefore  divide  the  angle  (=90°)  between  point  0 
and  6  into  six  equal  parts;  with  corresponding  points  on  the 
axis  of  beam  (see  Fig.  43)  and  note  that 

cos20°  cos2  15°  cos2  30°  cos2  90° 

UQ  =  —  j  —  ;  ui  =  —  j  -  ;  u2  =  —  j  --  ;  etc.,  to  UQ  =  —  j  --  . 


Hence,  eq.  (215)  becomes 

/V    2^p_    4  cos2  15°    2  cos2  30° 
E   •  + 


4  cos2  45°     2  cos2  60°     4  cos2  75°      1 
H T—  — j—  —j—    -+0    ;    .     (2K>> 

^3  *4  ^5 

or,  with  numerical  values,  using  the  inch  and  pound  as  units, 


l         4x0933Q     2X0.7500 
y  ~  28000000  X  18  L0.7854"f    0.7251          0.5692 

4X0.5000     2x0.2500     4x0.067 


0.06701 

Ti6ir.  =L05ln- 


as  the  value  of  the  vertical  deflection  0  ...  a. 

Similarly,  we  may  determine  the  horizontal  deflection 
a  ...  6',  using  the  same  method  as  in  §  64  (involving  a  " dummy  " 
horizontal  force  at  point  6). 

(It  is  here  supposed  that  the  elastic  limit  is  not  passed  in 
the  outer  fiber  of  any  section  of  the  curved  beam.  Let  the 
student  test  this  point). 


APPENDIX 


Integral  Forms.  A  few  integral  forms,  useful  in  the  mathe- 
matical work  of  this  book,  are  here  presented. 

/xn+1 
xndx= -• 
n+r 

/i'dx 
J   x 

J  sin  6  .  d6=  —cos  6; 

fcos  6  .  dd=  +sin#; 

J  sin  6  .  cos  0  .  dO  =  J  sin2  6\ 

\0  .  cos  0.  dd  =  6  .  sin  0  +  cos  0; 

c  • 

J  s 

fcos2  d  .  dO  =  \Q  +  i  sin  26. 

Note  C.  (See  foot  of  p.  30.)  If  a  curve  be  plotted  from 
the  expression  for  U,  as  function  of  TV  in  the  seventh  line 
from  top  of  p.  30,  with  U  as  (vertical)  ordinate  and  T\  as 
(horizontal)  abscissa,  this  curve  is  evidently  the  common  parabola; 
and,  moreover,  one  with  its  geometrical  axis  vertical;  since  it 
must  be  a  conic  section,  the  highest  power  of  either  co-ordinate 
being  the  second,  and  since  it  is  impossible  to  get  more  than  one 
value  of  U  for  any  assigned  value  of  TV  These  characteristics 

123 


124  APPENDIX 

show  that  the  curve  cannot  be  an  ellipse,  nor  an  hyperbola, 
nor  a  parabola  with  its  axis  other  than  vertical. 

For  the  angle  a,  which  a  tangent  line  at  any  point  of  this 

curve   makes  with   the   horizontal,  we   have  tan  a  =-7™-,  and 

al  i 

this  is  given  in  eq.  (45);  which  is  an  equation  of  the  first 
degree  as  respects  T^  so  that  only  one  value  of  T\  satisfies 

the  condition  -7777- =  0.     The  point  of  the  curve  where  -77^-  =  0 
a  1 1  al  i 

is,  of  course,  the  vertex  of  the  parabola,  where  the  tangent  line 
is  horizontal,  and  here  the  ordinate  U  is  either  a  maximum  or 
a  minimum.  It  will  be  a  minimum  if  the  curve  on  either  side 
of  the  vertex  rises,  the  final  value  of  U  being  +  infinity;  but  a 
maximum,  if  the  curve  descends  on  either  side,  to  a  final  ordinate 
of  —infinity.  But  the  latter  value  for  U  is  impossible;  since, 
from  the  nature  of  the  expression  for  U,  each  term  is  positive 
for  all  values  of  T\.  Hence  the  value  of  TI  obtained  by  putting 

-77--  =  0  corresponds  to  a  minimum  value  of  U. 

Note  D.  (See  p.  116.)  To  prove  eqs.  (199)  a  little  more 
specifically  let  us  suppose  that  the  rib  is  at  first  supported 
(i.e.,  held)  by  the  right-hand  support  at  K,  only,  and  under 
no  load;  but  at  the  higher  temperature,  t.  The  point  0  of  the 
axis  would  then  be  found  a  distance  A0,  horizontally  to  the  left 
from  where  it  is  actually  constrained  to  be,  and  the  plane  of 
the  cross-section  would  be  parallel  to  its  ultimate  position. 
Let  now  the  system  of  independent  "  loads,"  consisting  of  H 
and  the  forces  forming  the  couple  of  moment  M0,  be  gradually 
applied-  (beginning  with  zero  values)  until  they  reach  their 
respective  ultimate  and  actual  values.  The  horizontal  dis- 
placement of  point  0  of  the  rib  will  be  to,  =  y(t— t0)2r  cos  00, 

giving  -rjj  -  h  (p.  20) ;  while  the  angle  through  which  the  plane 
of  section  will  have  turned  is  zero,  whence  "7117"  =  0  (p.  25). 

.Q 


INDEX 


Angular  displacement  of  arm  of 
couple 24 

Arch  rib,  or  curved  beam 96 

Arch  rib,  segmental,  hinged  ends  106 

Arch  rib,  segmental,  no  hinges, 
fixed  ends. .. 109 

Arch  rib,  segmental,  tempera- 
ture stresses 109,  116 

Arch  rib,  semicircular,  hinged 
ends . 96 

Arch  rib,  semicircular,  no  hinges, 
fixed  ends.... 101 

Arch  rib,  semicircular,  with  tie 
rod 98 

Arch  rib,  semicircular,  tempera- 
ture stresses 100 

Bars  considered  elastic  in  succes- 
sion    1 

Bars,  necessary 2 

Bars,  redundant 2 

Beams,  curved 45,  94-122 

Beams  in  Flexure,  Internal  work  45 
Beams,  non-prismatic. .  .  .  60,  61,  65 
Beam  supported  by  parabolic 

.cable 86 

Beam  supported  by  three  others     78 

Beams,  straight 49-77 

Beam,  trussed 80 

Beam  with  overhang 53 

Brake-beam 80 

Cable  supporting  beam 86 

Cambered  trussed-beam 82 

Cambria  steel  I-beam 51 

Cantilever  beams 59 

Castigliano's  Theorem. 15,  17 

Castigliano's  Theorem,  geomet- 
rical proof 18 

Clapeyron's  Theorem 72 

Composite  systems 78 

Compressive  shortening 2 

Continuous  girders  ...   62,  64,  05-67, 

69-77 


Continuous  girder  of  variable 
moment  of  inertia 65 

Couple,  Derivative  of  vork  with 
respect  to  moment  of 24 

Curved  beams 45,  94-122 

Davit  of  uniform  section,  deflec- 
tion   94 

Davit  of  variable  cross-section, 
deflection 120 

Deflection  of  beams 49,  52,  53, 

59,60 

Deflection  of  beam  with  variable 
moment  of  inertia 53,  60,  61 

Derivatives  of  work 17,  20,  24 

Derivative  of  internal  work  as  to 
moment  of  an  external  couple .  24 

Displacement  of  joint  of  frame  . 

4,  20 

Displacement  of  point  in  axis  of 
beam 48 

Displacement,  variable,  Theo- 
rem for 18 

Double-knee  beam 90 

Elastic  limit  not  passed 1 

Elongation  of  bar 12,  13 

Equation  to  curve  found  by  Cas- 
tigliano's theorem 56,  59 

Example  of  deflection  of  joint  of 

frame 20 

External  work 4,  11 

Flexure,  work  of,  in  beams 45 

Fraenkel's  Formula 22,  53 

Frame  of  one  redundant  bar  .  .  5,  26 
Frame  of  two  redundant  bars. .  .     33 
Frame  of  more  than  two  redun- 
dant bars 37 

Gradual  application  of  loads  ...  1,  3 

Hiroi,  Prof.,  book  by 89 

Hudson,  C.  W.,  on  deflection  of 

beams 53 

125 


126 


INDKX 


I-beam,  work  of  shear  in 47,  51 

Initial  stresses 7,  11 

Integral  forms  (Appendix) 123 

Integrals  in  the  three-moment 

Theorem 72 

Internal  work 4,  13 

Internal  work  equals  external 

work. .  13 


Knee-loeam. 


90 

Least  work,  Principle  of 31 

Least  work,  a  particular  case  of 

more  general  relation 33 

Limitation  to  zero  initial  stresses 

11,  13 

Loads,  gradual  application  of . . .  1 

Martin,  H.  W.  book  by 89 

Mensch,  problem  treated  by. ...  93 
Moment,  internal,  statically  in- 
determinate, derivative  of  in- 
ternal work  with  respect  to. . .  68 
Moment     of     inertia,     variable, 

beams  of 60,  61,  65 

Moment,  "  normal " 71 

"  Necessary  "  bars 2 

"  Neutra)  "  force 18 

Non-prismatic  beams 60,61,65 

"  Normal  moment " 71 

Overhang,  beam  with 53 

Parabolic  cable  sustaining  beam 

and  load 86 

Projection  of  displacement 4 

"Redundant  bars". .  .   2,  5,  8,  11,  13 
Redundant  bar  too  short  or  too 

long 37 

Rough  supporting  surface 18 

Segmental     circular     arch     rib 

hinged  at  ends 106 


Segmental  circular  arch  rib  fixed 
at  ends,  no  hinges 109 

Shear,  work  of,  in  flexure.   47,  48,  51 

Shear,  influence  on  deflection  of 
I-beam 51 

Simpson's  Rule  for  approximate 
integration 118 

Simpson's  Rule  applied  in  prob- 
lem of  davit 120 

Smooth  supporting  surfaces  pos- 
tulated   18 

Statically  indeterminate  struc- 
tures   26 

Surfaces,  smooth 18 

Surfaces,  rough 18 

Tapering  beams 60,  61,  65 

Temperature 2,  13 

Temperature  stresses,  etc 2,  13, 

43,  84,  100,  109,  116 
Temperature  stresses  in  elastic 

bars 43 

Temperature  stresses  in  trussed- 

beam 84 

Theorem  of  three  moments 69 

Three  moments,  Theorem  of 69 

Thrust,  work  of 46 

Tilden,  Prof.,  article  by 51 

Trussed-beam 80 

Trussed-beam,  cambered 82 

Trussed  beam  with  two  struts 

and  three  tie-rods 84 

Variable  displacement  theorem  .     18 
Variable  moment  of  inertia  . .  .60,  61, 

65 
Virtual  velocities 3,  6,  7,  9 

Work  of  deformation -.  .  .       5 

Work,  external 4,12 

Work,  internal 4 

Work  of  shear  considered 51 

Work    of    shear    neglected     in 

nearly  all  cases 48 

Work  of  thrust  neglected. . .   105,  121 


Trigonometric   Ratios  (Natural);  including  "arc,"  by  which  is  meant 
•"radians,"  or  "7r-measure, "  or  "circular  measure ;"  e.g.,  arc  100°  =  1.7453293, 


arc 

de- 
gree 

sin     cosec 

tan    cot  an 

sec     cos 

0 

0 

I       in  fin. 

)       infin. 

1.0000  1.0000 

90 

1.5708 

0.0175 

1 

0.0175  57.299 

0.0175  57.290 

L0001  0.9998 

89 

1.5533 

.0349 

2 

.0349  28.654 

.0349  28.636 

1.0006   .9994 

88 

1.5359 

.0524 

3 

.0523  19.107 

.0524  19.081 

1.0014   .9986 

87 

1.5184 

.0698 

4 

.0698  14.336 

.0699  14.301 

1.0024   .9976 

86 

1.5010 

.0873 

5 

.0872  11.474 

.0875  11.430 

1.0038   .9962 

85 

1.4835 

0.1047 

6 

0.1045   9.5668 

0.1051   9.5144 

1.0055  0.9945 

84 

1.4661 

.1222 

7 

.1219   8.2055 

.1228   8.1443 

1.0075   .9925 

83 

1.4486 

.1396 

8 

.1392   7.1853 

.1405   7.1154 

1.0098   .9903 

82 

1.4312 

.1571 

9 

.1564   6.3925 

.1584   6.3138 

1.0125   .9877 

81 

1.4137 

.1745 

10 

.1736   5.7588 

.1763   5.6713 

1.0154   .9848 

80 

1.3963 

0.1920 

11 

0.1908   5.2408 

0.1944   5.1446 

1.0187  0.9816 

79 

1.3788 

.2094 

12 

.2079   4.8097 

.2126   4.7046 

1.0223   .9781 

78 

1.3614 

.2269 

13 

.2250   4.4454 

.2309   4.3315 

1.0263   .9744 

77 

1.3439 

.2443 

14 

.2419   4.1336 

.2493   4.0108 

1.0306   .9703 

76 

1.3264 

.2618 

15 

.2588   3.8637 

.2679   3.7321 

1.0353   .9659 

75 

1.3090 

0.2793 

16 

0.2756   3.6280 

0.2867   3.4874 

1.0403  0.9613 

74 

1.2915 

.2967 

17 

.2924   3.4203 

.3057   3.2709 

1.0457   .9563 

73 

1.2741 

.3142 

18 

.3090   3.2361 

.3249   3.0777 

1.0515   .9511 

72 

1.2566 

.3316 

19 

.3256   3.0716 

.3443   2.9042 

1.0576   .9455 

71 

1.2392 

.3491 

20 

.3420   2.9238 

.3640   2.7475 

1.0642   .9397 

70 

1.2217 

0.3665 

21 

0.3584   2.7904 

0.3839   2.6051 

1.0712  0.9336 

69 

1.2043 

.3840 

22 

.3746   2.6695 

.4040   2.4751 

1.0785   .9272 

68 

1.1868 

.4014 

23 

.3907   2.5593 

.4245   2.3559 

1.0864   .9205 

67 

1.1694 

.4189 

24 

.4067   2.4586 

.4452   2.2460 

1.0946   .9135 

66 

1.1519 

.4363 

25 

.4226   2.3662 

.4663   2.1445 

1.1034   .9063 

65 

1.1345 

0.4538 

26 

0.4384   2.2812 

0.4877   2.0503 

1.1126  0.8988 

64 

1.1170 

.4712 

27 

.4540   2.2027 

.5095   1.9626 

1.1223   .8910 

63 

1.0996 

.4887 

28 

.4695   2.1301 

.5317   1.8807 

1.1326   .8829 

62 

1.0821 

.5061 

29 

.4848   2.0627 

.5543   1.8040 

1.1434   .8746 

61 

1.0646 

.5236 

30 

.5000   2.0000 

.5774   1.7321 

1.1547   .8660 

60 

1.0472 

0.5411 

31 

0.5150   1.9416 

0.6009   1.6643 

1.1666  0.8572 

59 

1.0297 

.5585 

32 

.5299   1.8871 

.6249   1.6003 

1.1792   .8480 

58 

1.0123 

.5760 

33 

.5446   1.8361 

.6494   1.5399 

1.1924   .8387 

57 

0.9948 

.5934 

34 

.5592   1.7883 

.6745   1.4826 

1.2062   .8290 

56 

0.9774 

.6109 

35 

.5736   1.7435 

.7002   1.4281 

1.2208   .8192 

55 

0.9599 

0.6283 

36 

0.5878   1.7013 

0.7265   1.3764 

1.2361  0.8090 

54 

0.9425 

.6458 

37 

.6018   1.6616 

.7536   1.3270 

1.2521   .7986 

53 

0.9250 

.6632 

38 

.6157   1.6243 

.7813   1.2799 

1.2690   .7880 

52 

0.907G 

.6807 

39 

.6293   1.5890 

.8098   1.-2349 

1.2868   .7771 

51 

0.8901 

.6981 

40 

.6428   1.5557 

.8391   1.1918 

1.3054   .7660 

50 

0.8727 

0.7156 

41 

0.6561   1.5243 

0.8693   1.1504 

1.3250  0.7547 

49 

0.8552 

7330 

42 

.6691   1.4945 

.9004   1.1106 

1.3456   .7431 

48 

0.8378 

7505 

43 

.6820   1.4663 

.9325   1.0724 

1.3673   .7314 

47 

0.8203 

.7679 

44 

.6947   1.4396 

.9657   1  0355 

1.3902   .7193 

46 

0.8028 

.7854 

45 

.7071   1.4142 

1.0000   1  0000 

1.4142   .707.1 

45 

0.7854 

cos      sec 

c-otan     t  a  a 

cosec     sin 

de- 

arc 

gree 

OVERDUE. 


OCT 


DEC  V* 


tf* 

MAR  21    '946 
SEP  131946 


f  EB  ~0  1939 


DEC  6 


D 


oct 


MAR  2  1942 

1)» 


•  4  1954  LU 

- 
REC'D  LD 

MAR  19 


LD  21A-60m-4,'64 
(E4555slO)476B 


LD  2l- 


General  Library 

University  of  California 

Berkeley 


